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

Theorem

 * $\displaystyle \int \frac {x^2 \ \mathrm d x} {a x + b} = \frac {\left({a x + b}\right)^2} {2 a^3} - \frac {2 b \left({a x + b}\right)} {a^3} + \frac {b^2} {a^3} \ln \left\vert{a x + b}\right\vert + C$

Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
 * $\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a} - \frac {m b} {\left({m + n + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^n \ \mathrm d x$

Let $m = 2$ and $n = -1$.

Then: