Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x/Proof 1

Proof
Let $s \in \Z$.

Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.

Then:

Let $s$ be selected such that $m + n + 1 - s = 0$.

Then $s = m + n + 1$.

Thus $(2)$ after rearrangement becomes:
 * $\dfrac {\d u} {\d x} = \dfrac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}$

Then:

and:

Thus by Integration by Parts:
 * $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$