# Primitive of x over x cubed plus a cubed squared

## Theorem

$\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2} = \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

## Proof

First a lemma:

### Lemma

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

$\Box$

Then:

 $\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2}$ $=$ $\ds \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }$ from Lemma $\ds$ $=$ $\ds \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \paren {\frac 1 {6 a} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }$ Primitive of $\dfrac x {\paren {x^3 + a^3} }$ $\ds$ $=$ $\ds \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ simplifying

$\blacksquare$