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

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Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 - a^2}^2$

$\dfrac 1 {x^2 \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \paren {x + a} } - \dfrac 3 {4 a^5 \paren {x - a} } + \dfrac 1 {4 a^4 \paren {x + a}^2} + \dfrac 1 {4 a^4 \paren {x - a}^2}$


Proof

\(\ds \frac 1 {x^2 \paren {x^2 - a^2}^2}\) \(=\) \(\ds \frac 1 {x^2 \paren {x + a}^2 \paren {x - a}^2}\) Difference of Two Squares
\(\ds \) \(\equiv\) \(\ds \frac A {x + a} + \frac B {\paren {x + a}^2} + \frac C {x - a} + \frac D {\paren {x - a}^2} + \frac E x + \frac F {x^2}\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\equiv\) \(\ds A x^2 \paren {x^2 - a^2} \paren {x - a} + B x^2 \paren {x - a}^2\) multiplying through by $x^2 \paren {x^2 - a^2}^2$
\(\ds \) \(\) \(\, \ds + \, \) \(\ds C x^2 \paren {x^2 - a^2} \paren {x + a} + D x^2 \paren {x + a}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds E x \paren {x^2 - a^2}^2 + F \paren {x^2 - a^2}^2\)
\(\text {(1)}: \quad\) \(\ds \) \(\equiv\) \(\ds A x^5 - A a x^4 - A a^2 x^3 + A a^3 x^2 + B x^4 - 2 B a x^3 + B a^2 x^2\) multiplying out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds C x^5 + C a x^4 - C a^2 x^3 - C a^3 x^2 + D x^4 + 2 D a x^3 + D a^2 x^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds E x^5 - 2 E a^2 x^3 + E a^4 x + F x^4 - 2 F a^2 x^2 + F a^4\)


Setting $x = a$ in $(1)$:

\(\ds D a^2 \paren {2 a}^2\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds \frac 1 {4 a^4}\)


Setting $x = -a$ in $(1)$:

\(\ds B \paren {-a}^2 \paren {-2 a}^2\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds B\) \(=\) \(\ds \frac 1 {4 a^4}\)


Setting $x = 0$ in $(1)$:

\(\ds F a^4\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds F\) \(=\) \(\ds \frac 1 {a^4}\)


Equating coefficients of $x$ in $(1)$:

\(\ds 0\) \(=\) \(\ds E\)


Equating coefficients of $x^5$ in $(1)$:

\(\text {(2)}: \quad\) \(\ds 0\) \(=\) \(\ds A + C + E\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds A + C\) as $E = 0$
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds - C\)


Equating coefficients of $x^2$ in $(1)$:

\(\ds A a^3 + B a^2 - C a^3 + D a^2 - 2 F a^2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A a^3 + \frac 1 {4 a^4} a^2 - C a^3 + \frac 1 {4 a^4} a^2 - 2 \frac 1 {a^4} a^2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A + \frac 1 {4 a^5} - C + \frac 1 {4 a^5} - \frac 8 {4 a^5}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A - C\) \(=\) \(\ds \frac {8 - 1 - 1} {4 a^5}\)
\(\ds \) \(=\) \(\ds \frac 3 {2 a^5}\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds \frac 3 {4 a^5}\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds \frac {-3} {4 a^5}\) as $A = - C$


Summarising:

\(\ds A\) \(=\) \(\ds \frac 3 {4 a^5}\)
\(\ds B\) \(=\) \(\ds \frac 1 {4 a^4}\)
\(\ds C\) \(=\) \(\ds \frac {-3} {4 a^5}\)
\(\ds D\) \(=\) \(\ds \frac 1 {4 a^4}\)
\(\ds E\) \(=\) \(\ds 0\)
\(\ds F\) \(=\) \(\ds \frac 1 {a^4}\)


Hence the result.

$\blacksquare$

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