Primitive of Reciprocal of x squared plus a squared squared

Theorem

 * $\displaystyle \int \frac {\d x} {\paren {x^2 + a^2}^2} = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C$

Proof
Let:

Also see

 * Primitive of $\dfrac 1 {\paren {x^2 - a^2}^2}$
 * Primitive of $\dfrac 1 {\paren {a^2 - x^2}^2}$