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

Theorem

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

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:


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

Then:

and:

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