Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion 0 33732 407613 220626 2019-06-12T20:43:55Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)^2}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)^2} \equiv \dfrac {3 a^2} {b^4 x} - \dfrac {2 a} {b^3 x^2} + \dfrac 1 {b^2 x^3} - \dfrac {3 a^3} {b^4 \left({a x + b}\right)} - \dfrac {a^3} {b^3 \left({a x + b}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b} + \dfrac E {\left({a x + b}\right)^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right)^2 + B x \left({a x + b}\right)^2 + C \left({a x + b}\right)^2 + D x^3 \left({a x + b}\right) + E x^3       | c = multiplying through by $x^3 \left({a x + b}\right)^2$ }} {{eqn | o = \equiv       | r = A a^2 x^4 + 2 A a b x^3 + A b^2 x^2       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^3 + 2 B a b x^2 + B b^2 x       | c = (tedious though this is, it helps to }} {{eqn | o =        | ro= +       | r = C a^2 x^2 + 2 C a b x + C b^2       | c = identify the equal indices) }} {{eqn | o =        | ro= +       | r = D a x^4 + D b x^3 + E x^3       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = E \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = E       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b^2       | c =  }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 C a b + B b^2       | c =  }} {{eqn | ll= \implies       | l = \frac {2 a b} {b^2}       | r = -B b^2       | c = subtituting for $C$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = B       | r = -\frac {2 a} {b^3}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^2 + 2 B a b + C a^2       | c =  }} {{eqn | r = A b^2 + 2 \left({-\frac {2 a} {b^3} }\right) a b + \left({\frac 1 {b^2} }\right) a^2       | c = substituting for $B$ and $C$ from $(2)$ and $(3)$ }} {{eqn | ll= \implies       | l = A b^2       | r = \frac {4 a^2} {b^2} - \frac {a^2} {b^2}       | c = rearranging }} {{eqn | n = 4       | ll= \implies       | l = A       | r = \frac {3 a^2} {b^4}       | c = simplifying }} {{end-eqn}} Equating $4$th powers of $x$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + D a       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac {3 a^3} {b^4}       | c = substituting for $A$ from $(3)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {3 a^2} {b^4} }} {{eqn | l = B       | r = -\frac {2 a} {b^3} }} {{eqn | l = C       | r = \frac 1 {b^2} }} {{eqn | l = D       | r = \frac {3 a^3} {b^4} }} {{eqn | l = E       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata hk1k80vrcy5vr3kj71xhdbac62iz99x


Primitive of Reciprocal of x squared by a x + b squared/Partial Fraction Expansion 0 33733 407620 220627 2019-06-12T20:46:01Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)^2}$ == :$\dfrac 1 {x^2 \left({a x + b}\right)^2} \equiv -\dfrac {2 a} {b^3 x} + \dfrac 1 {b^2 x^2} + \dfrac {2 a^2} {b^3 \left({a x + b}\right)} + \dfrac {a^2} {b^2 \left({a x + b}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a x + b}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b} + \dfrac D {\left({a x + b}\right)^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a x + b}\right)^2 + B \left({a x + b}\right)^2 + C x^2 \left({a x + b}\right) + D x^2       | c = multiplying through by $x^2 \left({a x + b}\right)^2$ }} {{eqn | o = \equiv       | r = A a^2 x^3 + 2 A a b x^2 + A b^2 x       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^2 + 2 B a b x + B b^2       | c = (tedious though this is, it helps to }} {{eqn | o =        | ro= +       | r = C a x^3 + C b x^2 + D x^2       | c = identify the equal indices) }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = D \left({-\frac b a}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = D       | r = \frac {a^2} {b^2}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = B b^2       | c =  }} {{eqn | n = 2       | ll= \implies       | l = B       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^2 + 2 B a b       | c =  }} {{eqn | ll= \implies       | l = A b^2       | r = -\frac {2 a b} {b^2}       | c = subtituting for $B$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = A       | r = -\frac {2 a} {b^3}       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + C a       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {2 a^2} {b^3}       | c = substituting for $A$ from $(3)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac {2 a} {b^3} }} {{eqn | l = B       | r = \frac 1 {b^2} }} {{eqn | l = C       | r = \frac {2 a^2} {b^3} }} {{eqn | l = D       | r = \frac {a^2} {b^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata 7t9zwmc4yk753nn430cy797l47zy5ct


Primitive of Reciprocal of x by a x + b squared/Partial Fraction Expansion 0 33734 495810 407598 2020-10-21T20:53:17Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^2}$ == :$\dfrac 1 {x \paren {a x + b}^2} \equiv \dfrac 1 {b^2 x} - \dfrac a {b^2 \paren {a x + b} } - \dfrac a {b \paren {a x + b}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {a x + b}^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b} + \dfrac C {\paren {a x + b}^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a x + b}^2 + B x \paren {a x + b} + C x       | c = multiplying through by $x \paren {a x + b}^2$ }} {{eqn | o = \equiv       | r = A a^2 x^2 + 2 A a b x + A b^2 + B a x^2 + B b x + C x       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \leadsto       | l = C \paren {-\frac b a}       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \leadsto       | l = C       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b^2       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = A       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + B a       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac a {b^2}       | c = substituting for $A$ from $(2)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {b^2} }} {{eqn | l = B       | r = -\frac a {b^2} }} {{eqn | l = C       | r = -\frac a b }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata rg82b461fpapczoti63aa49viuxbwoj


Primitive of Reciprocal of x cubed by a x + b/Partial Fraction Expansion 0 33735 407614 220629 2019-06-12T20:44:13Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)} \equiv \dfrac {a^2} {b^3 x} + \dfrac {-a} {b^2 x^2} + \dfrac 1 {b x^3} + \dfrac {-a^3} {b^3 \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right) + B x \left({a x + b}\right) + C \left({a x + b}\right) + D x^3       | c = multiplying through by $x^3 \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x^3 + A b x^2 + B a x^2 + B b x + C a x + C b + D x^3       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = D \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | n = 2       | ll= \implies       | l = D       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac 1 b       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a + D       | c =  }} {{eqn | n = 3       | ll= \implies       | l = A       | r = \frac {a^2} {b^3}       | c = substituting for $D$ from $(2)$ and simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b + B a       | c =  }} {{eqn | r = \frac {a^2} {b^3} b + B a       | c = substituting for $A$ from $(3)$ }} {{eqn | ll= \implies       | l = B       | r = -\frac a {b^2}       | c = rearranging and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {a^2} {b^3} }} {{eqn | l = B       | r = -\frac a {b^2} }} {{eqn | l = C       | r = \frac 1 b }} {{eqn | l = D       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata f4ynmdmwy0z1el2feperqi60dv7rab3


Primitive of Reciprocal of x squared by a x + b/Partial Fraction Expansion 0 33736 407621 220630 2019-06-12T20:46:15Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)}$ == :$\dfrac 1 {x^2 \left({a x + b}\right)} \equiv -\dfrac a {b^2 x} + \dfrac 1 {b x^2} + \dfrac {a^2} {b^2 \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a x + b}\right) + B \left({a x + b}\right) + C x^2       | c = multiplying through by $x^2 \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x^2 + A b x + B a x + B b + C x^2       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = C \left({-\frac b a}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {a^2} {b^2}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = B b       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 b       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac a {b^2}       | c = substituting for $C$ from $(2)$ and rearranging }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac a {b^2} }} {{eqn | l = B       | r = \frac 1 b }} {{eqn | l = C       | r = \frac {a^2} {b^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata fagh8fgr6gihgoev0j5vk8a4t178et7


Primitive of Reciprocal of x by a x + b/Partial Fraction Expansion 0 33737 407599 220631 2019-06-12T20:38:04Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x \left({a x + b}\right)}$ == :$\dfrac 1 {x \left({a x + b}\right)} \equiv \dfrac 1 {b x} - \dfrac a {b \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a x + b}\right) + B x       | c = multiplying through by $x \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x + A b + B x       | c = multiplying out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({-\frac b a}\right)       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = B       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 b       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 b }} {{eqn | l = B       | r = -\frac a b }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata ker6cbgwr67q36i17hcd1890g5lvgt6


Primitive of Reciprocal of x by a x + b cubed/Partial Fraction Expansion 0 33745 407596 220653 2019-06-12T20:36:51Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^3}$ == :$\dfrac 1 {x \paren {a x + b}^3} \equiv \dfrac 1 {b^3 x} - \dfrac a {b^3 \paren {a x + b} } - \dfrac a {b^2 \paren {a x + b}^2} - \dfrac a {b \paren {a x + b}^3}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {a x + b}^3}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b} + \dfrac C {\paren {a x + b}^2} + \dfrac D {\paren {a x + b}^3}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a x + b}^3 + B x \paren {a x + b}^2 + C x \paren {a x + b} + D x       | c = multiplying through by $x \paren {a x + b}^3$ }} {{eqn | o = \equiv       | r = A a^3 x^3 + 3 A a^2 b x^2 + 3 A a b^2 x + A b^3       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^3 + 2 B a b x^2 + B b^2 x + C a x^2 + C b x + D x       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \leadsto       | l = D \paren {-\frac b a}       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \leadsto       | l = D       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b^3       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = A       | r = \frac 1 {b^3}       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^3 + B a^2       | c =  }} {{eqn | ll= \leadsto       | l = B a^2       | r = -\frac {a^3} {b^3}       | c = substituting for $A$ from $(2)$ }} {{eqn | n = 3       | ll= \leadsto       | l = B       | r = -\frac a {b^3}       | c = simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 3 A a^2 b + 2 B a b + C a       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = -3 A a b - 2 B b       | c = rerranging }} {{eqn | r = -3 a b \frac 1 {b^3} + 2 b \frac a {b^3}       | c = substituting for $A$ from $(2)$ and $B$ from $(3)$ }} {{eqn | r = -3 \frac a {b^2} + 2 \frac a {b^2}       | c = simplifying }} {{eqn | r = -\frac a {b^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {b^3} }} {{eqn | l = B       | r = -\frac a {b^3} }} {{eqn | l = C       | r = -\frac a {b^2} }} {{eqn | l = D       | r = -\frac a b }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata 9gna4h51gj3p4fkmtvxmmptx7xo65o7


Primitive of Reciprocal of x squared by a x + b cubed/Partial Fraction Expansion 0 33747 407618 220652 2019-06-12T20:45:31Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)^3}$ == :$\dfrac 1 {x^2 \left({a x + b}\right)^3} \equiv \dfrac {-3 a} {b^4 x} + \dfrac 1 {b^3 x^2} + \dfrac {3 a^2} {b^4 \left({a x + b}\right)} + \dfrac {2 a^2} {b^3 \left({a x + b}\right)^2} + \dfrac {a^2} {b^2 \left({a x + b}\right)^3}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a x + b}\right)^3}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b} + \dfrac D {\left({a x + b}\right)^2} + \dfrac E {\left({a x + b}\right)^3}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a x + b}\right)^3 + B \left({a x + b}\right)^3 + C x^2 \left({a x + b}\right)^2 + D x^2 \left({a x + b}\right) + E x^2       | c = multiplying through by $x^2 \left({a x + b}\right)^3$ }} {{eqn | o = \equiv       | r = A a^3 x^4 + 3 A a^2 b x^3 + 3 A a b^2 x^2 + A b^3 x       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^3 x^3 + 3 B a^2 b x^2 + 3 B a b^2 x + B b^3       | c =  }} {{eqn | o =        | ro= +       | r = C a^2 x^4 + 2 C a b x^3 + C b^2 x^2 + D a x^3 + D b x^2 + E x^2       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = E \left({-\frac b a}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = E       | r = \frac {a^2} {b^2}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = B b^3       | c =  }} {{eqn | n = 2       | ll= \implies       | l = B       | r = \frac 1 {b^3}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^3 + 3 B a b^2       | c =  }} {{eqn | ll= \implies       | l = A b^3       | r = -3 \frac 1 {b^3} a b^2       | c = substituting for $B$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = A       | r = \frac {-3 a} {b^4}       | c = simplifying }} {{end-eqn}} Equating $4$th powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^3 + C a^2       | c =  }} {{eqn | ll= \implies       | l = C a^2       | r = \frac {3 a} {b^4} a^3       | c = substituting for $A$ from $(3)$ }} {{eqn | n = 4       | ll= \implies       | l = C       | r = \frac {3 a^2} {b^4}       | c = simplifying }} {{end-eqn}} Equating $3$rd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 3 A a^2 b + B a^3 + 2 C a b + D a       | c =  }} {{eqn | ll= \implies       | l = D       | r = -3 A a b - B a^2 - 2 C b       | c = rearranging }} {{eqn | r = -3 a b \frac {-3 a} {b^4} - \frac 1 {b^3} a^2 - 2 b \frac {3 a^2} {b^4}       | c = substituting for $A$ from $(3)$, $B$ from $(2)$ and $C$ from $(4)$ }} {{eqn | r = \frac {9 a^2} {b^3} - \frac {a^2} {b^3} - \frac {6 a^2} {b^3}       | c = simplifying }} {{eqn | r = \frac {2 a^2} {b^3}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-3 a} {b^4} }} {{eqn | l = B       | r = \frac 1 {b^3} }} {{eqn | l = C       | r = \frac {3 a^2} {b^4} }} {{eqn | l = D       | r = \frac {2 a^2} {b^3} }} {{eqn | l = E       | r = \frac {a^2} {b^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata 8x5pzzoxvxf4n4dse5oqpjokx8vk2yr


Primitive of Reciprocal of x cubed by a x + b cubed/Partial Fraction Expansion 0 33755 407611 220619 2019-06-12T20:43:25Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)^3}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)^3} \equiv \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \left({a x + b}\right)} -\dfrac {3 a^3} {b^4 \left({a x + b}\right)^2} - \dfrac {a^3} {b^3 \left({a x + b}\right)^3}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)^3}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b} + \dfrac E {\left({a x + b}\right)^2} + \dfrac F {\left({a x + b}\right)^3}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right)^3 + B x \left({a x + b}\right)^3 + C \left({a x + b}\right)^3 + D x^3 \left({a x + b}\right)^2 + E x^2 \left({a x + b}\right) + F x^2       | c = multiplying through by $x^2 \left({a x + b}\right)^3$ }} {{eqn | o = \equiv       | r = A a^3 x^5 + 3 A a^2 b x^4 + 3 A a b^2 x^3 + A b^3 x^2       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^3 x^4 + 3 B a^2 b x^3 + 3 B a b^2 x^2 + B b^3 x       | c =  }} {{eqn | o =        | ro= +       | r = C a^3 x^3 + 3 C a^2 b x^2 + 3 C a b^2 x + C b^3       | c =  }} {{eqn | o =        | ro= +       | r = D a^2 x^5 + 2 D a b x^4 + D b^2 x^3 + E a x^4 + E b x^3 + F x^3       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = F \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = F       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b^3       | c =  }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac 1 {b^3}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B b^3 + 3 C a b^2       | c =  }} {{eqn | ll= \implies       | l = B b^3       | r = -3 \frac 1 {b^3} a b^2       | c = substituting for $C$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = B       | r = \frac {-3 a} {b^4}       | c = simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^3 + 3 B a b^2 + 3 C a^2 b       | c =  }} {{eqn | ll= \implies       | l = A b^3       | r = -3 \frac {-3 a} {b^4} a b^2 - 3 \frac 1 {b^3} a^2 b       | c = substituting for $C$ from $(2)$ and $B$ from $(3)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {9 a^2} {b^5} - 3 \frac {a^2} {b^5}       | c = simplifying }} {{eqn | n = 4       | r = \frac {6 a^2} {b^5}       | c = simplifying }} {{end-eqn}} Equating $5$th powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^3 + D a^2       | c =  }} {{eqn | ll= \implies       | l = D a^2       | r = -\frac {6 a^2} {b^5} a^3       | c = substituting for $A$ from $(4)$ }} {{eqn | n = 5       | ll= \implies       | l = D       | r = -\frac {6 a^3} {b^5}       | c = simplifying }} {{end-eqn}} Equating $4$th powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 3 A a^2 b + B a^3 + 2 D a b + E a       | c =  }} {{eqn | ll= \implies       | l = E       | r = -3 A a b - B a^2 - 2 D b       | c = rearranging }} {{eqn | r = -3 a b \frac {6 a^2} {b^5} - \frac {-3 a} {b^4} a^2 - 2 \frac {-6 a^3} {b^5} b       | c = substituting for $A$ from $(4)$, $B$ from $(3)$ and $D$ from $(5)$ }} {{eqn | r = -\frac {18 a^3} {b^4} + \frac {3 a^3} {b^4} + \frac {12 a^3} {b^4}       | c = simplifying }} {{eqn | r = -\frac {3 a^3} {b^4}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {6 a^2} {b^5} }} {{eqn | l = B       | r = -\frac {3 a} {b^4} }} {{eqn | l = C       | r = \frac 1 {b^3} }} {{eqn | l = D       | r = -\frac {6 a^3} {b^5} }} {{eqn | l = E       | r = -\frac {3 a^3} {b^4} }} {{eqn | l = F       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b/Lemmata t9cz268t2m64gf2f1xex6swv2eu44e9


Primitive of Root of a x + b over Power of x/Formulation 1 0 33874 188090 2014-05-24T06:10:32Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m -..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m - 1}\right)} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$ </onlyinclude> == Proof == Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u}{\mathrm d x}       | r = \frac a {2 \sqrt{a x + b} }       | c = Power Rule for Derivatives etc. }} {{eqn | l = v       | r = \frac {-1} {\left({m - 1}\right) x^{m - 1} }       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d v}{\mathrm d x}       | r = \frac 1 {x^m}       | c = Power Rule for Derivatives }} {{end-eqn}} From Integration by Parts: : $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$ from which: {{begin-eqn}} {{eqn | l = \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x       | r = \int \sqrt{a x + b} \frac 1 {x^m} \ \mathrm d x       | c =  }} {{eqn | r = \sqrt{a x + b} \frac {-1} {\left({m - 1}\right) x^{m - 1} } - \int \frac {-1} {\left({m - 1}\right) x^{m - 1} } \frac a {2 \sqrt{a x + b} } \ \mathrm d x       | c =  }} {{eqn | r = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m - 1}\right)} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Power of x by Root of a x + b|next=Primitive of Root of a x + b over Power of x/Formulation 2}}: $\S 14$: Integrals involving $\sqrt{a x + b}$: $14.97$ Category:Primitives involving Root of a x + b pk9gve2i0ecocasv4ubaegm50wsvp19


Primitive of Reciprocal of a x + b by p x + q/Partial Fraction Expansion 0 33886 188115 188113 2014-05-24T19:15:15Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of a x + b by p x + q == :$\dfrac 1 {\left({a x + b}\right) \left({p x + q}\right)} \equiv \dfrac {-a} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac p {\left({b p - a q}\right) \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({a x + b}\right) \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({p x + q}\right) + B \left({a x + b}\right)       | c = multiplying through by $\left({a x + b}\right) \left({p x + q}\right)$ }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = B \left({a \left({-\frac q p}\right) + b}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $p x + q$ is $0$ }} {{eqn | ll= \implies       | l = B \left({\frac {b p - a q} p}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac p {b p - a q}       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = A \left({p \left({-\frac b a}\right) + q}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = A \left({\frac {-b p + a q} a}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-a} {b p - a q}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-a} {p b - a q} }} {{eqn | l = B       | r = \frac p {b p - a q} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b and p x + q/Lemmata 5971lztczzaln1zy4hpeeazsdflryak


Primitive of x over a x + b by p x + q/Partial Fraction Expansion 0 33889 191849 188120 2014-08-19T22:16:01Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of x over a x + b by p x + q == :$\dfrac x {\left({a x + b}\right) \left({p x + q}\right)} \equiv \dfrac b {\left({b p - a q}\right) \left({a x + b}\right)} - \dfrac q {\left({b p - a q}\right) \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac x {\left({a x + b}\right) \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = x       | o = \equiv       | r = A \left({p x + q}\right) + B \left({a x + b}\right)       | c = multiplying through by $\left({a x + b}\right) \left({p x + q}\right)$ }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = B \left({a \left({-\frac q p}\right) + b}\right)       | r = -\frac q p       | c = substituting for $x$ in $(1)$: term in $p x + q$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-q} {b p - a q}       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = A \left({p \left({-\frac b a}\right) + q}\right)       | r = -\frac b a       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = A       | r = \frac b {b p - a q}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac b {p b - a q} }} {{eqn | l = B       | r = \frac {-q} {b p - a q} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b and p x + q/Lemmata 1u26x3kg8tj7eax9vec04ldj16pfvot


Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion 0 33891 188129 188126 2014-05-24T21:28:47Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of a x + b squared by p x + q == :$\dfrac 1 {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {-a p} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac {-a} {\left({a x + b}\right)^2} + \dfrac {p^2} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({a x + b}\right)^2 \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {\left({a x + b}\right)^2} + \frac C {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a x + b}\right) \left({p x + q}\right) + B \left({p x + q}\right) + C \left({a x + b}\right)^2       | c = multiplying through by $\left({a x + b}\right)^2 \left({p x + q}\right)$ }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({p \left({-\frac b a}\right) + q}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-a} {b p - a q}       | c =  }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = C \left({a \left({-\frac q p}\right) + b}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {p^2} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C \frac a p       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac {p^2} {\left({b p - a q}\right)^2} \frac a p       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {-a p} {\left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-a p} {\left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac {-a} {b p - a q} }} {{eqn | l = C       | r = \frac {p^2} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b and p x + q/Lemmata t465uyeblrnmm6m8k04s7fdd575abd6


Primitive of x over a x + b squared by p x + q/Partial Fraction Expansion 0 33893 188130 2014-05-24T21:29:54Z Prime.mover 59 Created page with "== Lemma for Primitive of x over a x + b squared by p x + q ==  :$\dfrac x {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {a q..." wikitext text/x-wiki == Lemma for Primitive of x over a x + b squared by p x + q == :$\dfrac x {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {a q} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac b {\left({a x + b}\right)^2} - \dfrac {p q} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)$ == Proof == {{begin-eqn}} {{eqn | l = \frac x {\left({a x + b}\right)^2 \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {\left({a x + b}\right)^2} + \frac C {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = x       | o = \equiv       | r = A \left({a x + b}\right) \left({p x + q}\right) + B \left({p x + q}\right) + C \left({a x + b}\right)^2       | c = multiplying through by $\left({a x + b}\right)^2 \left({p x + q}\right)$ }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({p \left({-\frac b a}\right) + q}\right)       | r = -\frac b a       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac b {b p - a q}       | c =  }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = C \left({a \left({-\frac q p}\right) + b}\right)^2       | r = -\frac q p       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {-p q} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C \frac a p       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac {p q} {\left({b p - a q}\right)^2} \frac a p       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {a q} {\left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {a q} {\left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac b {b p - a q} }} {{eqn | l = C       | r = \frac {-p q} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b and p x + q/Lemmata 1fsrcgxs6v6e1zjqvc301kuubv2lk1b


Primitive of x squared over a x + b squared by p x + q/Partial Fraction Expansion 0 33896 221069 219845 2015-06-12T06:48:52Z Kc kennylau 2331 wikitext text/x-wiki == Lemma for Primitive of $\dfrac {x^2} {\left({a x + b}\right)^2 \left({p x + q}\right)}$ == :$\dfrac {x^2} {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2 \left({a x + b}\right)} + \dfrac {-b^2} {a \left({b p - a q}\right) \left({a x + b}\right)^2} + \dfrac {q^2} {\left({b p - a q}\right)^2 \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac {x^2} {\left({a x + b}\right)^2 \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {\left({a x + b}\right)^2} + \frac C {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = x^2       | o = \equiv       | r = A \left({a x + b}\right) \left({p x + q}\right) + B \left({p x + q}\right) + C \left({a x + b}\right)^2       | c = multiplying through by $\left({a x + b}\right)^2 \left({p x + q}\right)$ }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({p \left({-\frac b a}\right) + q}\right)       | r = \frac {b^2} {a^2}       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-b^2} {a \left({b p - a q}\right)}       | c =  }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = C \left({a \left({-\frac q p}\right) + b}\right)^2       | r = \frac {q^2} {p^2}       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {q^2} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A a p       | r = 1 - \frac {q^2} {\left({b p - a q}\right)^2} a^2       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a p} - \frac {a q^2} {p \left({b p - a q}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {\left({b p - a q}\right)^2 - a^2 q^2} {a p \left({b p - a q}\right)^2}       | c = common denominator }} {{eqn | ll= \implies       | l = A       | r = \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac {-b^2} {a \left({b p - a q}\right)} }} {{eqn | l = C       | r = \frac {q^2} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a x + b and p x + q/Lemmata 48bk2wpymjpu6yecpmawqt7e6w9pq00


Primitive of Reciprocal of x by x squared plus a squared/Partial Fraction Expansion 0 33949 454169 188334 2020-03-12T23:23:24Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \paren {x^2 + a^2}$ == :$\dfrac 1 {x \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x} - \dfrac x {a^2 \paren {x^2 + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {x^2 + a^2} }       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {x^2 + a^2} + B x^2 + C x       | c = multiplying through by $x \paren {x^2 + a^2}$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^2} }} {{eqn | l = B       | r = -\frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata 4c3dqw2x7sri1z1phrcjhev095ulb8w


Primitive of Reciprocal of x squared by x squared plus a squared/Partial Fraction Expansion 0 33952 188332 2014-05-30T17:22:09Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 + a^2}\right)$ ==  :$\dfrac 1 {x^2 \left({x^2 + a^..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({x^2 + a^2}\right)$ == :$\dfrac 1 {x^2 \left({x^2 + a^2}\right)} \equiv \dfrac 1 {a^2 x^2} - \dfrac 1 {a^2 \left({x^2 + a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({x^2 + a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({x^2 + a^2}\right) + B \left({x^2 + a^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({x^2 + a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata fvx9pp1l8t2jnv8imw5utyyvu26gvf6


Primitive of Reciprocal of x cubed by x squared plus a squared/Partial Fraction Expansion 0 33954 391500 188336 2019-02-05T14:13:36Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 + a^2}$ == :$\dfrac 1 {x^3 \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \paren {x^2 + a^2} }       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A x^2 \paren {x^2 + a^2} + B x \paren {x^2 + a^2} + C \paren {x^2 + a^2} + D x^4 + E x^3       | c = multiplying through by $x^3 \paren {x^2 + a^2}$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = C a^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B a^2       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + C       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \leadsto       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac 1 {a^4} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac 1 {a^2} }} {{eqn | l = D       | r = \frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata 5wrkm4i3eax5r7zwncgbtftiu335t7m


Primitive of Reciprocal of x by x squared plus a squared squared/Partial Fraction Expansion 0 33960 188356 2014-05-31T06:41:14Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x \left({x^2 + a^2}\right)^2$ ==  :$\dfrac 1 {x \left({x^2 + a^2}..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({x^2 + a^2}\right)^2$ == :$\dfrac 1 {x \left({x^2 + a^2}\right)^2} \equiv \dfrac 1 {a^4 x} - \dfrac x {a^4 \left({x^2 + a^2}\right)} - \dfrac x {a^2 \left({x^2 + a^2}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({x^2 + a^2}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 + a^2} + \dfrac {D x + E} {\left({x^2 + a^2}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 + a^2}\right)^2 + \left({B x + C}\right) x \left({x^2 + a^2}\right) + \left({D x + E}\right) x       | c = multiplying through by $x \left({x^2 + a^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A x^4 + 2 A a^2 x^2 + A a^4 + B x^4 + B x^2 a^2 + C x^3 + C x a^2 + D x^2 + E x       | c = multiplying everything out }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 A a^2 + B a^2 + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^4} }} {{eqn | l = B       | r = -\frac 1 {a^4} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^2} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata f4x9gicqomtsttw7s3vq6c7g1e6yhgq


Primitive of Reciprocal of x squared by x squared plus a squared squared/Partial Fraction Expansion 0 33962 437864 188358 2019-12-05T17:50:19Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 + a^2}^2$ == :$\dfrac 1 {x^2 \paren {x^2 + a^2}^2} \equiv -\dfrac 1 {a^4 x^2} - \dfrac 1 {a^4 \paren {x^2 + a^2} } - \dfrac 1 {a^2 \paren {x^2 + a^2}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \paren {x^2 + a^2}^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 + a^2} + \dfrac {E x + F} {\paren {x^2 + a^2}^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x \paren {x^2 + a^2}^2 + B \paren {x^2 + a^2}^2       | c = multiplying through by $x^2 \paren {x^2 + a^2}^2$ }} {{eqn | o =       | ro= +       | r = \paren {C x + D} x^2 \paren {x^2 + a^2} + \paren {E x + F} x^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^5 + 2 A a^2 x^3 + A a^4 x + B x^4 + 2 B a^2 x^2 + B a^4       | c = multiplying everything out }} {{eqn | o =       | ro= +       | r = C x^5 + C x^3 a^2 + D x^4 + D x^2 a^2 + E x^3 + F x^2       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 2 A a^2 + C a^2 + E       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 B a^2 + D a^2 + F       | c =  }} {{eqn | ll= \leadsto       | l = F       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^4} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = -\frac 1 {a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata i0rxwsr3cu7dj83xtmmjzhct4fu84vg


Primitive of Reciprocal of x cubed by x squared plus a squared squared/Partial Fraction Expansion 0 33964 188381 188367 2014-05-31T19:18:11Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 \left({x^2 + a^2}\right)^2$ == :$\dfrac 1 {x^3 \left({x^2 + a^2}\right)^2} \equiv -\dfrac 2 {a^6 x} + \dfrac 1 {a^4 x^3} + \dfrac {2 x} {a^6 \left({x^2 + a^2}\right)} + \dfrac x {a^4 \left({x^2 + a^2}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({x^2 + a^2}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 + a^2} + \dfrac {F x + G} {\left({x^2 + a^2}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 + a^2}\right)^2 + B x \left({x^2 + a^2}\right)^2 + C \left({x^2 + a^2}\right)^2       | c = multiplying through by $x^2 \left({x^2 + a^2}\right)^2$ }} {{eqn | o =       | ro= +       | r = \left({D x + E}\right) x^3 \left({x^2 + a^2}\right) + \left({F x + G}\right) x^3       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^6 + 2 A a^2 x^4 + A a^4 x^2 + B x^5 + 2 B a^2 x^3 + B a^4 x       | c = multiplying everything out }} {{eqn | o =       | ro= +       | r = C x^4 + 2 C a^2 x^2 + C a^4       | c =  }} {{eqn | o =       | ro= +       | r = D x^6 + D x^4 a^2 + E x^5 + E x^3 a^2 + F x^4 + G x^3       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = C a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^4 + 2 C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^6$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 B a^2 + E a^2 + G       | c =  }} {{eqn | ll= \implies       | l = G       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 2 A a^2 + C + D a^2 + F       | r = 0       | c =  }} {{eqn | ll= \implies       | l = F       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac 2 {a^6} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac 1 {a^4} }} {{eqn | l = D       | r = \frac 2 {a^6} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = \frac 1 {a^4} }} {{eqn | l = G       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared plus a squared/Lemmata 6v2lbonc5pgtp69cwguk6924m9en808


Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Proof 2/Partial Fraction Expansion 0 33971 188509 188406 2014-06-03T21:43:18Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $\left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^2 - a^2} \equiv \dfrac 1 {2 a \left({x - a}\right)} - \dfrac 1 {2 a \left({x + a}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({x^2 - a^2}\right)}       | o = \equiv       | r = \frac A {x - a} + \frac B {x + a}       | c = Difference of Two Squares }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x + a}\right) + B \left({x - a}\right)       | c = multiplying through by $\left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 2 a + B \cdot 0       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 0 + B \cdot \left({-2 a}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac {-1} {2 a}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a} }} {{eqn | l = B       | r = \frac {-1} {2 a} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata na99tpvc4gybnvq4kqp4y4zcir52ex4


Primitive of Reciprocal of x by x squared minus a squared/Partial Fraction Expansion 0 33977 188429 188428 2014-06-01T20:23:54Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x \left({x^2 - a^2}\right)} \equiv \dfrac {-1} {a^2 x} + \dfrac x {a^2 \left({x^2 - a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 - a^2}\right) + B x^2 + C x       | c = multiplying through by $x \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-1} {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^2} }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata ns5f3xz1rqc7sjq55xthrwl7hjjtweb


Primitive of Reciprocal of x squared by x squared minus a squared/Partial Fraction Expansion 0 33979 188698 188431 2014-06-06T19:41:21Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^2 \left({x^2 - a^2}\right)} \equiv \dfrac 1 {a^2 \left({x^2 - a^2}\right)} - \dfrac 1 {a^2 x^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({x^2 - a^2}\right) + B \left({x^2 - a^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac {-1} {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = \frac 1 {a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata 2ee6aeecv76b3a7by34d04guom9jqy6


Primitive of Reciprocal of x cubed by x squared minus a squared/Partial Fraction Expansion 0 33981 188702 188434 2014-06-06T21:12:33Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^3 \left({x^2 - a^2}\right)} \equiv \dfrac {-1} {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \left({x^2 - a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 - a^2}\right) + B x \left({x^2 - a^2}\right) + C \left({x^2 - a^2}\right) + D x^4 + E x^3       | c = multiplying through by $x^3 \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -C a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {-1} {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -B a^2       | c =  }} {{eqn | ll= \implies       | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A a^2 + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-1} {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^4} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac {-1} {a^2} }} {{eqn | l = D       | r = \frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata c1gjx0v8dmhhjpcpuago13il6qci1py


Primitive of Reciprocal of x squared minus a squared squared/Partial Fraction Expansion 0 33983 188711 188708 2014-06-06T22:01:13Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $\left({x^2 - a^2}\right)^2$ == :$\dfrac 1 {\left({x^2 - a^2}\right)^2} \equiv \dfrac 1 {4 a^3 \left({x + a}\right)} - \dfrac 1 {4 a^3 \left({x - a}\right)} + \dfrac 1 {4 a^2 \left({x + a}\right)^2} + \dfrac 1 {4 a^2 \left({x - a}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({x^2 - a^2}\right)^2}       | r = \dfrac 1 {\left({x + a}\right)^2 \left({x - a}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \dfrac A {x + a} + \dfrac B {\left({x + a}\right)^2} + \dfrac C {x - a} + \dfrac D {\left({x - a}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 - a^2}\right) \left({x - a}\right) + B \left({x - a}\right)^2 + C \left({x^2 - a^2}\right) \left({x + a}\right) + D \left({x + a}\right)^2       | c = multiplying through by $\left({x^2 - a^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A x^3 - A a x^2 - A a^2 x + A a^3 + B x^2 - 2 B a x + B a^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^3 + C a x^2 - C a^2 x - C a^3 + D x^2 + 2 D a x + D a^2       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({-2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a + C a + B + D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = - A a + C a + \frac 1 {4 a^2} + \frac 1 {4 a^2}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = C a + C a + \frac 1 {2 a^2}       | r = 0       | c = as $A = - C$ }} {{eqn | ll= \implies       | l = 2 C        | r = \frac {-1} {2 a^3}       | c =  }} {{eqn | ll= \implies       | l = C        | r = \frac {-1} {4 a^3}       | c =  }} {{eqn | ll= \implies       | l = A        | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {4 a^3} }} {{eqn | l = B       | r = \frac 1 {4 a^2} }} {{eqn | l = C       | r = \frac {-1} {4 a^3} }} {{eqn | l = D       | r = \frac 1 {4 a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata nr708mewttsrky3erlm3hpj1w4hjf3i


Primitive of Reciprocal of x by x squared minus a squared squared/Partial Fraction Expansion 0 34003 429199 188727 2019-10-02T12:57:03Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \paren {x^2 - a^2}^2$ == :$\dfrac 1 {x \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x} + \dfrac {-x} {a^4 \paren {x^2 - a^2} } + \dfrac x {a^2 \paren {x^2 - a^2}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x \paren {x^2 - a^2}^2}       | r = \frac 1 {x \paren {x + a}^2 \paren {x - a}^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac B {\paren {x + a}^2} + \frac C {x - a} + \frac D {\paren {x - a}^2} + \frac E x       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x \paren {x^2 - a^2} \paren {x - a} + B x \paren {x - a}^2       | c = multiplying through by $x \paren {x^2 - a^2}^2$ }} {{eqn | o =        | ro= +       | r = C x \paren {x^2 - a^2} \paren {x + a} + D x \paren {x + a}^2 + E \paren {x^2 - a^2}^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^4 - A a x^3 - A a^2 x^2 + A a^3 x + B x^3 - 2 B a x^2 + B a^2 x       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^4 + C a x^3 - C a^2 x^2 - C a^3 x + D x^3 + 2 D a x^2 + D a^2 x       | c =  }} {{eqn | o =        | ro= +       | r = E x^4 - 2 E a^2 x^2 + E a^4       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \paren {-a} \paren {-2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac {-1} {4 a^3}       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = E a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = E       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C + E       | c =  }} {{eqn | ll= \leadsto       | r = C       | l = -A - \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a + C a + B + D       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = - A a + C a + \frac 1 {4 a^3} + \frac {-1} {4 a^3}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -A a + \paren {-A - \frac 1 {a^4} } a       | r = 0       | c = as $C = -A  -\dfrac 1 {a^4}$ }} {{eqn | ll= \leadsto       | l = -2 A - \frac 1 {a^4}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A        | r = \frac {-1} {2 a^4}       | c =  }} {{eqn | ll= \leadsto       | l = C        | r = \frac {-1} {2 a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {2 a^4} }} {{eqn | l = B       | r = \frac {-1} {4 a^3} }} {{eqn | l = C       | r = \frac {-1} {2 a^4} }} {{eqn | l = D       | r = \frac 1 {4 a^3} }} {{eqn | l = E       | r = \frac 1 {a^4} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x \paren {x^2 - a^2}^2}       | o = \equiv       | r = \frac {-1} {2 a^4 \paren {x + a} } + \frac {-1} {4 a^3 \paren {x + a}^2} + \frac {-1} {2 a^4 \paren {x - a} } + \frac 1 {4 a^3 \paren {x - a}^2} + \frac 1 {a^4 x}       | c =  }} {{eqn | o = \equiv       | r = - \frac {\paren {x - a} + \paren {x + a} } {2 a^4 \paren {x + a} \paren {x - a} } + \frac {\paren {x + a}^2 - \paren {x - a}^2} {4 a^3 \paren {x + a}^2 \paren {x - a}^2} + \frac 1 {a^4 x}       | c = common denominators }} {{eqn | o = \equiv       | r = \frac {- 2 x} {2 a^4 \paren {x^2 - a^2} } + \frac {\paren {x^2 + 2 a x + a^2} - \paren {x^2 - 2 a x + a^2} } {4 a^3 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac {-x} {a^4 \paren {x^2 - a^2} } + \frac {4 a x} {4 a^3 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \dfrac 1 {a^4 x} + \dfrac {-x} {a^4 \paren {x^2 - a^2} } + \dfrac x {a^2 \paren {x^2 - a^2}^2}       | c = simplifying and rearranging }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata aeeicmwtpjlf71202sglewj9w43ev11


Primitive of Reciprocal of x squared by x squared minus a squared squared/Partial Fraction Expansion 0 34005 188492 2014-06-03T15:13:21Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)^2$ ==  :$\dfrac 1 {x^2 \lef..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)^2$ == :$\dfrac 1 {x^2 \left({x^2 - a^2}\right)^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \left({x + a}\right)} - \dfrac 3 {4 a^5 \left({x - a}\right)} + \dfrac 1 {4 a^4 \left({x + a}\right)^2} + \dfrac 1 {4 a^4 \left({x - a}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^2 \left({x^2 - a^2}\right)^2}       | r = \frac 1 {x^2 \left({x + a}\right)^2 \left({x - a}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac B {\left({x + a}\right)^2} + \frac C {x - a} + \frac D {\left({x - a}\right)^2} + \frac E x + \frac F {x^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 - a^2}\right) \left({x - a}\right) + B x^2 \left({x - a}\right)^2       | c = multiplying through by $x^2 \left({x^2 - a^2}\right)^2$ }} {{eqn | o =        | ro= +       | r = C x^2 \left({x^2 - a^2}\right) \left({x + a}\right) + D x^2 \left({x + a}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = E x \left({x^2 - a^2}\right)^2 + F \left({x^2 - a^2}\right)^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^5 - A a x^4 - A a^2 x^3 + A a^3 x^2 + B x^4 - 2 B a x^3 + B a^2 x^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^5 + C a x^4 - C a^2 x^3 - C a^3 x^2 + D x^4 + 2 D a x^3 + D a^2 x^2       | c =  }} {{eqn | o =        | ro= +       | r = E x^5 - 2 E a^2 x^3 + E a^4 x + F x^4 - 2 F a^2 x^2 + F a^4       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a^2 \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({-a}\right)^2 \left({-2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = F a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = F       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = E       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = 0       | r = A + C + E       | c =  }} {{eqn | ll= \implies       | l = 0       | r = A + C       | c = as $E = 0$ }} {{eqn | ll= \implies       | l = A       | r = - C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^3 + B a^2 - C a^3 + D a^2 - 2 F a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A a^3 + \frac 1 {4 a^4} a^2 - C a^3 + \frac 1 {4 a^4} a^2 - 2 \frac 1 {a^4} a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A + \frac 1 {4 a^5} - C + \frac 1 {4 a^5} - \frac 8 {4 a^5}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A - C       | r = \frac {8 - 1 - 1} {4 a^5}       | c =  }} {{eqn | r = \frac 3 {2 a^5}       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 3 {4 a^5}       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {-3} {4 a^5}       | c = as $A = - C$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 3 {4 a^5} }} {{eqn | l = B       | r = \frac 1 {4 a^4} }} {{eqn | l = C       | r = \frac {-3} {4 a^5} }} {{eqn | l = D       | r = \frac 1 {4 a^4} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = \frac 1 {a^4} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata 6kkxsyez5vos4oeben020jniion3cbp


Primitive of Reciprocal of x cubed by x squared minus a squared squared/Partial Fraction Expansion 0 34006 442084 188493 2020-01-02T18:07:27Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}^2$ == :$\dfrac 1 {x^3 \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac {2 x} {a^6 \paren {x^2 - a^2} } + \dfrac x {a^4 \paren {x^2 - a^2}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^3 \paren {x^2 - a^2}^2}       | r = \frac 1 {x^3 \paren {x + a}^2 \paren {x - a}^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac B {\paren {x + a}^2} + \frac C {x - a} + \frac D {\paren {x - a}^2} + \frac E x + \frac F {x^2} + \frac G {x^3}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x^3 \paren {x^2 - a^2} \paren {x - a} + B x^3 \paren {x - a}^2       | c = multiplying through by $x^3 \paren {x^2 - a^2}^2$ }} {{eqn | o =        | ro= +       | r = C x^3 \paren {x^2 - a^2} \paren {x + a} + D x^3 \paren {x + a}^2       | c =  }} {{eqn | o =        | ro= +       | r = E x^2 \paren {x^2 - a^2}^2 + F x \paren {x^2 - a^2}^2 + G \paren {x^2 - a^2}^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^6 - A a x^5 - A a^2 x^4 + A a^3 x^3 + B x^5 - 2 B a x^4 + B a^2 x^3       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^6 + C a x^5 - C a^2 x^4 - C a^3 x^3 + D x^5 + 2 D a x^4 + D a^2 x^3       | c =  }} {{eqn | o =        | ro= +       | r = E x^6 - 2 E a^2 x^4 + E a^4 x^2 + F x^5 - 2 F a^2 x^3 + F x a^4       | c =  }} {{eqn | o =        | ro= +       | r = G x^4 - 2 G a^2 x^2 + G a^4       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = G a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = G       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a^3 \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {4 a^5}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \paren {-a}^3 \paren {-2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac {-1} {4 a^5}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = F       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = E a^4 - 2 G a^2        | c =  }} {{eqn | ll= \leadsto       | l = E a^4       | r = 2 \frac 1 {a^4} a^2        | c =  }} {{eqn | ll= \leadsto       | l = E       | r = \frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^6$ in $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = 0       | r = A + C + E       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = A + C + \frac 2 {a^6}       | c =  }} {{eqn | ll= \leadsto       | l = A + C       | r = -\frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^3 + B a^2 - C a^3 + D a^2 - 2 F a^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A a^3 + \frac {-1} {4 a^5} a^2 - C a^3 + \frac 1 {4 a^5} a^2       | r = 0       | c = substituting for $B$, $D$ and $F$ }} {{eqn | ll= \leadsto       | l = A - C       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = C       | c =  }} {{eqn | r = \frac {-1} {a^6}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^6} }} {{eqn | l = B       | r = \frac {-1} {4 a^5} }} {{eqn | l = C       | r = \frac {-1} {a^6} }} {{eqn | l = D       | r = \frac 1 {4 a^5} }} {{eqn | l = E       | r = \frac 2 {a^6} }} {{eqn | l = F       | r = 0 }} {{eqn | l = G       | r = \frac 1 {a^4} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x^3 \paren {x^2 - a^2}^2}       | o = \equiv       | r = \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac 1 {a^6 \paren {x + a} } - \dfrac 1 {a^6 \paren {x - a} } - \dfrac 1 {4 a^5 \paren {x + a}^2} + \dfrac 1 {4 a^5 \paren {x - a}^2}       | c =  }} {{eqn | o = \equiv       | r = -\frac {\paren {x - a} + \paren {x + a} } {a^6 \paren {x + a} \paren {x - a} } + \frac {\paren {x + a}^2 - \paren {x - a}^2} {4 a^5 \paren {x + a}^2 \paren {x - a}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}       | c = common denominators }} {{eqn | o = \equiv       | r = \frac {- 2 x} {a^6 \paren {x^2 - a^2} } + \frac {\paren {x^2 + 2 a x + a^2} - \paren {x^2 - 2 a x + a^2} } {4 a^5 \paren {x^2 - a^2}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac {-2 x} {a^6 \paren {x^2 - a^2} } + \frac {4 a x} {4 a^5 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x^3} + \frac 2 {a^6 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac 1 {a^4 x^3} + \frac 2 {a^6 x} - \frac {2 x} {a^6 \paren {x^2 - a^2} } + \frac x {a^4 \paren {x^2 - a^2}^2}       | c = simplifying and rearranging }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x squared minus a squared/Lemmata qozn8wt7hcbtz4e5sddnxwxqvrri9r2


Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Partial Fraction Expansion 0 34015 491278 491233 2020-09-27T20:40:40Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $\dfrac 1 {a^2 - x^2}$ == :$\dfrac 1 {a^2 - x^2} \equiv \dfrac 1 {2 a \paren {a + x} } + \dfrac 1 {2 a \paren {a - x} }$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\paren {a^2 - x^2} }       | o = \equiv       | r = \frac A {a - x} + \frac B {a + x}       | c = Difference of Two Squares }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a + x} + B \paren {a - x}       | c = multiplying through by $\paren {a^2 - x^2}$ }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 2 a + B \cdot 0       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 0 + B \cdot \paren {2 a}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a} }} {{eqn | l = B       | r = \frac 1 {2 a} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata d534fuov29yxzvvm6akh93eluv35zge


Primitive of Reciprocal/Corollary

Primitive of Reciprocal/Corollary 0 34019 407574 220559 2019-06-12T19:52:11Z Prime.mover 59 wikitext text/x-wiki == Corollary to Primitive of Reciprocal == <onlyinclude> :$\displaystyle \int \frac {\d x} x = \ln x + C$ for $x > 0$. </onlyinclude> == Proof == From Primitive of Reciprocal: :$\displaystyle \int \frac {\d x} x = \ln \size x + C$ for $x \ne 0$. By definition of absolute value: :$\forall x \in \R_{>0}: \size x = x$ Hence the result. {{qed}} Category:Primitive of Reciprocal 2jrck878i0wl0as8k9diho1rzp3s0ai


Primitive of Reciprocal of x by a squared minus x squared/Partial Fraction Expansion 0 34052 188687 2014-06-06T18:57:30Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x \left({a^2 - x^2}\right)$ ==  :$\dfrac 1 {x \left({a^2 - x^2}\right)} ..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({a^2 - x^2}\right)$ == :$\dfrac 1 {x \left({a^2 - x^2}\right)} \equiv \dfrac 1 {a^2 x} + \dfrac x {a^2 \left({a^2 - x^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({a^2 - x^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {a^2 - x^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a^2 - x^2}\right) + B x^2 + C x       | c = multiplying through by $x \left({a^2 - x^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^2} }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata svznimzdo5r7p2mkavnrwderha3qx8k


Primitive of Reciprocal of x by a squared minus x squared 0 34053 188690 2014-06-06T19:03:04Z Prime.mover 59 Created page with "<onlyinclude> :$\displaystyle \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)} = \frac 1 {2 a^2} \ln \left({\frac {x^2} {a^2 - x^2} }\right) + C$ for $x^2 < a^2$. </onlyi..." wikitext text/x-wiki <onlyinclude> :$\displaystyle \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)} = \frac 1 {2 a^2} \ln \left({\frac {x^2} {a^2 - x^2} }\right) + C$ for $x^2 < a^2$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)}       | r = \int \left({\frac 1 {a^2 x} + \frac x {a^2 \left({a^2 - x^2}\right)} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} x + \frac 1 {a^2} \int \frac {x \ \mathrm d x} {a^2 - x^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{x}\right\vert + \frac 1 {a^2} \int \frac {x \ \mathrm d x} {a^2 - x^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{x}\right\vert + \frac 1 {a^2} \left({-\frac 1 2 \ln \left ({a^2 - x^2}\right) }\right) + C       | c = Primitive of $\dfrac x {a^2 - x^2}$ }} {{eqn | r = \frac 1 {2 a^2} \ln \left\vert{x^2}\right\vert - \frac 1 {2 a^2} \ln \left ({a^2 - x^2}\right) + C       | c = Logarithm of Power }} {{eqn | r = \frac 1 {2 a^2} \ln \left({x^2}\right) - \frac 1 {2 a^2} \ln \left ({a^2 - x^2}\right) + C       | c = as $x^2 > 0$ }} {{eqn | r = \frac 1 {2 a^2} \ln \left({\frac {x^2} {x^2 - a^2} }\right) + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\dfrac 1 {x \left({x^2 - a^2}\right)}$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of x cubed over a squared minus x squared|next=Primitive of Reciprocal of x squared by a squared minus x squared}}: $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.167$ Category:Primitives involving a squared minus x squared oknobe5myf7a8jvth8l65ngmd1nmvvg


Primitive of Reciprocal of x squared by a squared minus x squared/Partial Fraction Expansion 0 34054 188699 2014-06-06T19:42:33Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({a^2 - x^2}\right)$ ==  :$\dfrac 1 {x^2 \left({a^2 - x..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({a^2 - x^2}\right)$ == :$\dfrac 1 {x^2 \left({a^2 - x^2}\right)} \equiv \dfrac 1 {a^2 \left({a^2 - x^2}\right)} + \dfrac 1 {a^2 x^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a^2 - x^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {a^2 - x^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a^2 - x^2}\right) + B \left({a^2 - x^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({a^2 - x^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = \frac 1 {a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata 8i2q2a06gzxgi1c2b8y5yr13jl8leog


Primitive of Reciprocal of a squared minus x squared squared/Partial Fraction Expansion 0 34059 188714 188712 2014-06-06T22:11:34Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $\left({a^2 - x^2}\right)^2$ == :$\dfrac 1 {\left({a^2 - x^2}\right)^2} \equiv \dfrac 1 {4 a^3 \left({a + x}\right)} + \dfrac 1 {4 a^3 \left({a - x}\right)} + \dfrac 1 {4 a^2 \left({a + x}\right)^2} + \dfrac 1 {4 a^2 \left({a - x}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({a^2 - x^2}\right)^2}       | r = \dfrac 1 {\left({a + x}\right)^2 \left({a - x}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \dfrac A {a + x} + \dfrac B {\left({a + x}\right)^2} + \dfrac C {a - x} + \dfrac D {\left({a - x}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a^2 - x^2}\right) \left({a - x}\right) + B \left({a - x}\right)^2 + C \left({a^2 - x^2}\right) \left({a + x}\right) + D \left({a + x}\right)^2       | c = multiplying through by $\left({a^2 - x^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A a^3 - A a^2 x - A a x^2 + A x^3 + B a^2 - 2 B a x + B x^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C a^3 + C a^2 x - C a x^2 - C x^3 + D a^2 + 2 D a x + D x^2       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A - C       | c =  }} {{eqn | ll= \implies       | l = A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a - C a + B + D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = - A a - C a + \frac 1 {4 a^2} + \frac 1 {4 a^2}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -C a - C a + \frac 1 {2 a^2}       | r = 0       | c = as $A = C$ }} {{eqn | ll= \implies       | l = 2 C        | r = \frac 1 {2 a^3}       | c =  }} {{eqn | ll= \implies       | l = C        | r = \frac 1 {4 a^3}       | c =  }} {{eqn | ll= \implies       | l = A        | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {4 a^3} }} {{eqn | l = B       | r = \frac 1 {4 a^2} }} {{eqn | l = C       | r = \frac 1 {4 a^3} }} {{eqn | l = D       | r = \frac 1 {4 a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata trze76y5ilafa2hh7z3bpu3so6s4wb9


Primitive of Reciprocal of a squared minus x squared squared 0 34060 188713 2014-06-06T22:10:40Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^2} = \frac x {2 a^2 \left({a^2 - x^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^2} = \frac x {2 a^2 \left({a^2 - x^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + C$ for $x^2 < a^2$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^2}       | r = \int \left({\frac 1 {4 a^3 \left({a + x}\right)} + \frac 1 {4 a^3 \left({a - x}\right)} + \frac 1 {4 a^2 \left({a + x}\right)^2} + \frac 1 {4 a^2 \left({a - x}\right)^2} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {4 a^3} \int \frac {\mathrm d x} {a + x} + \frac 1 {4 a^3} \int \frac {\mathrm d x} {a - x} + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a + x}\right)^2} + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a - x}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {4 a^3} \ln \left\vert{a + x}\right\vert - \frac 1 {4 a^3} \ln \left\vert{a - x}\right\vert + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a + x}\right)^2} + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a - x}\right)^2} + C       | c = Primitive of Function of $a x + b$ and Primitive of Reciprocal }} {{eqn | r = \frac 1 {4 a^3} \ln \left\vert{\frac {a + x} {a - x} }\right\vert + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a + x}\right)^2} + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a - x}\right)^2} + C       | c = Difference of Logarithms }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a + x}\right)^2} + \frac 1 {4 a^2} \int \frac {\mathrm d x} {\left({a - x}\right)^2} + C       | c = Sign of Quotient of Factors of Difference of Squares }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + \frac 1 {4 a^2} \frac {-1} {\left({a + x}\right)} + \frac 1 {4 a^2} \frac 1 {\left({a - x}\right)} + C       | c = Primitive of Function of $a x + b$ and Primitive of Power }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + \left({\frac 1 {4 a^2 \left({a - x}\right)} - \frac 1  {4 a^2 \left({a + x}\right)} }\right) + C       | c = simplifying }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + \left({\frac {\left({a + x}\right) - \left({a - x}\right)} {4 a^2 \left({a - x}\right) \left({a + x}\right)} }\right) + C       | c = simplifying }} {{eqn | r = \frac x {2 a^2 \left({a^2 - x^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {a + x} {a - x} }\right) + C       | c = Difference of Two Squares }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\dfrac 1 {\left({x^2 - a^2}\right)^2}$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Reciprocal of x cubed by a squared minus x squared|next=Primitive of x over a squared minus x squared squared}}: $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.170$ Category:Primitives involving a squared minus x squared a35z20n0qx2m1syyf00x7pdmwa8p5pm


Primitive of Reciprocal of x by a squared minus x squared squared/Partial Fraction Expansion 0 34064 188730 188728 2014-06-07T12:50:03Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({a^2 - x^2}\right)^2$ == :$\dfrac 1 {x \left({a^2 - x^2}\right)^2} \equiv \dfrac 1 {a^4 x} + \dfrac x {a^4 \left({a^2 - x^2}\right)} + \dfrac x {a^2 \left({a^2 - x^2}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x \left({a^2 - x^2}\right)^2}       | r = \frac 1 {x \left({a + x}\right)^2 \left({a - x}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {a + x} + \frac B {\left({a + x}\right)^2} + \frac C {a - x} + \frac D {\left({a - x}\right)^2} + \frac E x       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a^2 - x^2}\right) \left({a - x}\right) + B x \left({a - x}\right)^2       | c = multiplying through by $x \left({a^2 - x^2}\right)^2$ }} {{eqn | o =        | ro= +       | r = C x \left({a^2 - x^2}\right) \left({a + x}\right) + D x \left({a + x}\right)^2 + E \left({a^2 - x^2}\right)^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^4 - A a x^3 - A a^2 x^2 + A a^3 x + B x^3 - 2 B a x^2 + B a^2 x       | c = multiplying out }} {{eqn | o =        | ro= -       | r = C x^4 - C a x^3 + C a^2 x^2 + C a^3 x + D x^3 + 2 D a x^2 + D a^2 x       | c =  }} {{eqn | o =        | ro= +       | r = E x^4 - 2 E a^2 x^2 + E a^4       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({-a}\right) \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac {-1} {4 a^3}       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = E a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = E       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A - C + E       | c =  }} {{eqn | ll= \implies       | r = C       | l = A + \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a - C a + B + D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = - A a - C a + \frac 1 {4 a^3} + \frac {-1} {4 a^3}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -A a - \left({A + \frac 1 {a^4} }\right) a       | r = 0       | c = as $C = A + \dfrac 1 {a^4}$ }} {{eqn | ll= \implies       | l = -2 A - \frac 1 {a^4}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A        | r = \frac {-1} {2 a^4}       | c =  }} {{eqn | ll= \implies       | l = C        | r = \frac 1 {2 a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {2 a^4} }} {{eqn | l = B       | r = \frac {-1} {4 a^3} }} {{eqn | l = C       | r = \frac 1 {2 a^4} }} {{eqn | l = D       | r = \frac 1 {4 a^3} }} {{eqn | l = E       | r = \frac 1 {a^4} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x \left({a^2 - x^2}\right)^2}       | o = \equiv       | r = \frac {-1} {2 a^4 \left({a + x}\right)} + \frac {-1} {4 a^3 \left({a + x}\right)^2} + \frac 1 {2 a^4 \left({a - x}\right)} + \frac 1 {4 a^3 \left({a - x}\right)^2} + \frac 1 {a^4 x}       | c =  }} {{eqn | o = \equiv       | r = \frac {\left({a + x}\right) - \left({a - x}\right)} {2 a^4 \left({a - x}\right) \left({a + x}\right)} + \frac {\left({a + x}\right)^2 - \left({a - x}\right)^2} {4 a^3 \left({a + x}\right)^2 \left({a - x}\right)^2} + \frac 1 {a^4 x}       | c = common denominators }} {{eqn | o = \equiv       | r = \frac {2 x} {2 a^4 \left({a^2 - x^2}\right)} + \frac {\left({a^2 + 2 a x + x^2}\right) - \left({a^2 - 2 a x + x^2}\right)} {4 a^3 \left({a^2 - x^2}\right)^2} + \frac 1 {a^4 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac x {a^4 \left({a^2 - x^2}\right)} + \frac {4 a x} {4 a^3 \left({a^2 - x^2}\right)^2} + \frac 1 {a^4 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \dfrac 1 {a^4 x} + \dfrac x {a^4 \left({a^2 - x^2}\right)} + \dfrac x {a^2 \left({a^2 - x^2}\right)^2}       | c = simplifying and rearranging }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata eyarnq5typ38bc2c0fs9p2k1b5ri6b9


Primitive of Reciprocal of x squared by a squared minus x squared squared/Partial Fraction Expansion 0 34066 476805 188735 2020-07-03T16:25:35Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \paren {a^2 - x^2}^2$ == :$\dfrac 1 {x^2 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \paren {a + x} } - \dfrac 3 {4 a^5 \paren {a - x} } + \dfrac 1 {4 a^4 \paren {a + x}^2} + \dfrac 1 {4 a^4 \paren {a - x}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^2 \paren {a^2 - x^2}^2}       | r = \frac 1 {x^2 \paren {a + x}^2 \paren {a - x}^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {a + x} + \frac B {\paren {a + x}^2} + \frac C {a - x} + \frac D {\paren {a - x}^2} + \frac E x + \frac F {x^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x^2 \paren {a^2 - x^2} \paren {a - x} + B x^2 \paren {a - x}^2       | c = multiplying through by $x^2 \paren {a^2 - x^2}^2$ }} {{eqn | o =        | ro= +       | r = C x^2 \paren {a^2 - x^2} \paren {a + x} + D x^2 \paren {a + x}^2       | c =  }} {{eqn | o =        | ro= +       | r = E x \paren {a^2 - x^2}^2 + F \paren {a^2 - x^2}^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^5 - A a x^4 - A a^2 x^3 + A a^3 x^2 + B x^4 - 2 B a x^3 + B a^2 x^2       | c = multiplying out }} {{eqn | o =        | ro= -       | r = C x^5 - C a x^4 + C a^2 x^3 + C a^3 x^2 + D x^4 + 2 D a x^3 + D a^2 x^2       | c =  }} {{eqn | o =        | ro= +       | r = E x^5 - 2 E a^2 x^3 + E a^4 x + F x^4 - 2 F a^2 x^2 + F a^4       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a^2 \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \paren {-a}^2 \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = F a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = F       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = E       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = 0       | r = A - C + E       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = A - C       | c = as $E = 0$ }} {{eqn | ll= \leadsto       | l = A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^3 + B a^2 + C a^3 + D a^2 - 2 F a^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A a^3 + \frac 1 {4 a^4} a^2 + C a^3 + \frac 1 {4 a^4} a^2 - 2 \frac 1 {a^4} a^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A + \frac 1 {4 a^5} + C + \frac 1 {4 a^5} - \frac 8 {4 a^5}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A + C       | r = \frac {8 - 1 - 1} {4 a^5}       | c =  }} {{eqn | r = \frac 3 {2 a^5}       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac 3 {4 a^5}       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = \frac 3 {4 a^5}       | c = as $A = C$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 3 {4 a^5} }} {{eqn | l = B       | r = \frac 1 {4 a^4} }} {{eqn | l = C       | r = \frac 3 {4 a^5} }} {{eqn | l = D       | r = \frac 1 {4 a^4} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = \frac 1 {a^4} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata om4bfn722d26r1bdgs42ux9vdreolbk