Primitive of Reciprocal of x cubed by x squared plus a squared squared

Theorem

 * $\displaystyle \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2} = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C$

Proof
Let: