Primitive of x squared over x cubed plus a cubed

Theorem

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

$\ds \map {\frac \d {\d x} } {x^3 + a^3}$

$=$

$\ds 3 x^2$

$\ds \leadsto \ \ $

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

$=$

$\ds \frac 1 3 \ln \size {x^3 + a^3} + C$

$\blacksquare$

$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$

So:

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

$=$

$\ds \frac 1 3 \ln \size {x^3 + a^3} + C$

directly.

$\blacksquare$