Primitive of Reciprocal of x cubed plus a cubed/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of $x^3 + a^3$

 * $\dfrac 1 {x^3 + a^3} = \dfrac 1 {3 a^2 \paren {x + a} } - \dfrac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} }$

Proof
Equating coefficients of $x^2$ in $(1)$:

Equating coefficients of $x$ in $(1)$:

Setting $x = 0$ in $(1)$:

Summarising:

Hence the result.