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


 * $\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}$

Then:

and:

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