Primitive of x over a x + b cubed/Proof 2

Theorem

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

Proof

From Primitive of $x$ by Power of $a x + b$:

$\ds \int x \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 2} } {\paren {n + 2} a^2} - \frac {b \paren {a x + b}^{n + 1} } {\paren {n + 1} a^2} + C$

where $n \ne - 1$ and $n \ne - 2$.

The result follows by setting $n = -3$.

$\blacksquare$