Primitive of Reciprocal of x cubed plus a cubed

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x^3 + a^3} = \frac 1 {6 a^2} \ln \left\vert{\frac {\left({x + a}\right)^2} {x^2 - a x + a^2} }\right\vert + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$