Primitive of x over x cubed plus a cubed squared/Lemma

Primitive of x over x cubed plus a cubed squared: Lemma

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

Proof
Then from Primitive of $\dfrac {\d x} {\paren {a x + b}^m \paren {p x + q}^n}$:
 * $\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$

Here we have $a = 1, b = 0, m = \dfrac 1 3, p = 1, q = a^3, n = 2$.

So: