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

## Theorem

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

## Proof 1

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

Substituting $m + 1$ for $m$:

 $\displaystyle \int x^{m + 1} \left({a x + b}\right)^n \rd x$ $=$ $\displaystyle \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \frac {\left({m + 1}\right) b} {\left({m + n + 2}\right) a} \int x^m \left({a x + b}\right)^n \rd x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\left({m + 1}\right) b} {\left({m + n + 2}\right) a} \int x^m \left({a x + b}\right)^n \rd x$ $=$ $\displaystyle \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \int x^{m + 1} \left({a x + b}\right)^n \rd x$ rearranging $\displaystyle \implies \ \$ $\displaystyle \int x^m \left({a x + b}\right)^n \rd x$ $=$ $\displaystyle \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$ rearranging $\displaystyle$ $=$ $\displaystyle \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$ rearranging

$\blacksquare$

## Proof 2

$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \rd x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{n + 1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n + 1} \rd x}\right)$

Setting $p := 1, q := 0, m := n, n := m$:

 $\displaystyle \int x^m \left({a x + b}\right)^n \rd x$ $=$ $\displaystyle \frac 1 {\left({m + 1}\right) \left({b 1 - a 0}\right)} \left({\left({a x + b}\right)^{n + 1} \left({1 x + 0}\right)^{m + 1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^n \left({0 x + 1}\right)^{m + 1} \rd x}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\left({m + 1}\right) b} \left({\left({a x + b}\right)^{n + 1} x^{m+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^n x^{m + 1} \rd x}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$

$\blacksquare$