Primitive of Reciprocal of x cubed plus a cubed

Theorem

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