Primitive of Reciprocal of x squared by x fourth minus a fourth

Theorem

 * $\ds \int \frac {\d x} {x^2 \paren {x^4 - a^4} } = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a^5} \arctan \frac x a + C$