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

Theorem

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

Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:


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

Substituting $n + 1$ for $n$: