Primitive of Reciprocal of x by x squared plus 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

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

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

 $\ds A a^4$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds A$ $=$ $\ds \frac 1 {a^4}$

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

 $\ds 0$ $=$ $\ds A + B$ $\ds \leadsto \ \$ $\ds B$ $=$ $\ds -\frac 1 {a^4}$

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

 $\ds C$ $=$ $\ds 0$

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

 $\ds 0$ $=$ $\ds C + E$ $\ds \leadsto \ \$ $\ds E$ $=$ $\ds 0$

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

 $\ds 0$ $=$ $\ds 2 A a^2 + B a^2 + D$ $\ds \leadsto \ \$ $\ds D$ $=$ $\ds -\frac 1 {a^2}$

Summarising:

 $\ds A$ $=$ $\ds \frac 1 {a^4}$ $\ds B$ $=$ $\ds -\frac 1 {a^4}$ $\ds C$ $=$ $\ds 0$ $\ds D$ $=$ $\ds -\frac 1 {a^2}$ $\ds E$ $=$ $\ds 0$

Hence the result.

$\blacksquare$