Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b/Proof 2

Theorem

 * $\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {-x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) b} + \frac {m + n + 2} {\left({n + 1}\right) b} \int x^m \left({a x + b}\right)^{n + 1} \ \mathrm d x$

Proof
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$

Setting $a := 1, b := 0, p x + q := a x + b$: