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

Theorem

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

Proof
Now consider: