Primitive of Reciprocal of x cubed by x squared plus a squared squared/Partial Fraction Expansion

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Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 + a^2}^2$

$\dfrac 1 {x^3 \paren {x^2 + a^2}^2} \equiv -\dfrac 2 {a^6 x} + \dfrac 1 {a^4 x^3} + \dfrac {2 x} {a^6 \paren {x^2 + a^2} } + \dfrac x {a^4 \paren {x^2 + a^2}^2}$


Proof

\(\ds \dfrac 1 {x^3 \paren {x^2 + a^2}^2}\) \(\equiv\) \(\ds \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 + a^2} + \dfrac {F x + G} {\paren {x^2 + a^2}^2}\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\equiv\) \(\ds A x^2 \paren {x^2 + a^2}^2 + B x \paren {x^2 + a^2}^2 + C \paren {x^2 + a^2}^2\) multiplying through by $x^2 \paren {x^2 + a^2}^2$
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {D x + E} x^3 \paren {x^2 + a^2} + \paren {F x + G} x^3\)
\(\text {(1)}: \quad\) \(\ds \) \(\equiv\) \(\ds A x^6 + 2 A a^2 x^4 + A a^4 x^2 + B x^5 + 2 B a^2 x^3 + B a^4 x\) multiplying everything out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds C x^4 + 2 C a^2 x^2 + C a^4\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds D x^6 + D x^4 a^2 + E x^5 + E x^3 a^2 + F x^4 + G x^3\)


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

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


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

\(\ds 0\) \(=\) \(\ds A a^4 + 2 C a^2\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds -\frac 2 {a^6}\)


Equating coefficients of $x^6$ in $(1)$:

\(\ds 0\) \(=\) \(\ds A + D\)
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds \frac 2 {a^6}\)


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

\(\ds B\) \(=\) \(\ds 0\)


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

\(\ds 0\) \(=\) \(\ds B + E\)
\(\ds \leadsto \ \ \) \(\ds E\) \(=\) \(\ds 0\)


Equating coefficients of $x^3$ in $(1)$:

\(\ds 0\) \(=\) \(\ds 2 B a^2 + E a^2 + G\)
\(\ds \leadsto \ \ \) \(\ds G\) \(=\) \(\ds 0\)


Equating coefficients of $x^4$ in $(1)$:

\(\ds 2 A a^2 + C + D a^2 + F\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds F\) \(=\) \(\ds \frac 1 {a^4}\)


Summarising:

\(\ds A\) \(=\) \(\ds -\frac 2 {a^6}\)
\(\ds B\) \(=\) \(\ds 0\)
\(\ds C\) \(=\) \(\ds \frac 1 {a^4}\)
\(\ds D\) \(=\) \(\ds \frac 2 {a^6}\)
\(\ds E\) \(=\) \(\ds 0\)
\(\ds F\) \(=\) \(\ds \frac 1 {a^4}\)
\(\ds G\) \(=\) \(\ds 0\)

Hence the result.

$\blacksquare$