Primitive of Reciprocal of x by x cubed plus a cubed

Theorem

 * $\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$

Proof
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
 * $\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$

Setting $n = 3$:


 * $\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$

directly.