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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)^2} = \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \left({x^3 + a^3}\right)} - \frac 4 {3 a^6} \int \frac {x \ \mathrm d x} {x^3 + a^3}$

Proof
Now consider: