Primitive of Reciprocal of x by x squared minus a squared squared/Partial Fraction Expansion
== 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.