Primitive of x squared over x cubed plus a cubed/Proof 2

Theorem

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

Proof

From Primitive of Power of x less one over Power of x plus Power of a:

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

So:

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

\(=\)

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

Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 3$

directly.

$\blacksquare$