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

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

\(\ds \dfrac 1 {x^2 \paren {x^2 + a^2}^2}\)

\(\equiv\)

\(\ds \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 + a^2} + \dfrac {E x + F} {\paren {x^2 + a^2}^2}\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A x \paren {x^2 + a^2}^2 + B \paren {x^2 + a^2}^2\)

multiplying through by $x^2 \paren {x^2 + a^2}^2$

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \paren {C x + D} x^2 \paren {x^2 + a^2} + \paren {E x + F} x^2\)

\(\text {(1)}: \quad\)

\(\ds \)

\(\equiv\)

\(\ds A x^5 + 2 A a^2 x^3 + A a^4 x + B x^4 + 2 B a^2 x^2 + B a^4\)

multiplying everything out

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds C x^5 + C x^3 a^2 + D x^4 + D x^2 a^2 + E x^3 + F x^2\)

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

\(\ds B a^4\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac 1 {a^4}\)

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

\(\ds 0\)

\(=\)

\(\ds B + D\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds -\frac 1 {a^4}\)

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

\(\ds A\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds A + C\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds 0\)

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

\(\ds 2 A a^2 + C a^2 + E\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds 2 B a^2 + D a^2 + F\)

\(\ds \leadsto \ \ \)

\(\ds F\)

\(=\)

\(\ds -\frac 1 {a^2}\)

Summarising:

\(\ds A\)

\(=\)

\(\ds 0\)

\(\ds B\)

\(=\)

\(\ds \frac 1 {a^4}\)

\(\ds C\)

\(=\)

\(\ds 0\)

\(\ds D\)

\(=\)

\(\ds -\frac 1 {a^4}\)

\(\ds E\)

\(=\)

\(\ds 0\)

\(\ds F\)

\(=\)

\(\ds -\frac 1 {a^2}\)

Hence the result.

$\blacksquare$