Primitive of x squared over x cubed plus a cubed squared

Theorem

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

$\ds z$

$=$

$\ds x^3 + a^3$

$\ds \leadsto \ \ $

$\ds \frac {\d z} {\d x}$

$=$

$\ds 3 x^2$

$\ds \leadsto \ \ $

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

$=$

$\ds \int \frac {\d z} {3 z^2}$

$\ds $

$=$

$\ds \frac {-1} {3 z} + C$

$\ds $

$=$

$\ds \frac {-1} {3 \paren {x^3 + a^3} } + C$

substituting for $z$

$\blacksquare$