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

Theorem

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

Proof
Let $s \in \Z$.

Let $u \dfrac {\mathrm d v} {\mathrm d x} = x^m \left({a x + b}\right)^n$.

Then:

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

Then $s = m + n + 1$.

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

Then:

and:

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