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

Theorem

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

Proof
Let: