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

Proof
From Reduction Formula for Primitive of $x^m \paren {a x + b}^n$: Decrement of Power of $x$:
 * $\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$

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

Then: