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