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

Proof
From Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: 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: