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

Theorem

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