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

Theorem

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