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

## Theorem

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

## Proof 1

Let $s \in \Z$.

 $\displaystyle v$ $=$ $\displaystyle \left({a x + b}\right)^s$ $(1):\quad$ $\displaystyle \implies \ \$ $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle a s \left({a x + b}\right)^{s - 1}$ Power Rule for Derivatives and Derivatives of Function of $a x + b$

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

Then:

 $\displaystyle u$ $=$ $\displaystyle \frac {x^m \left({a x + b}\right)^n} {a s \left({a x + b}\right)^{s - 1} }$ from $(1)$ $\displaystyle$ $=$ $\displaystyle \frac {x^m} {a s} \left({a x + b}\right)^{n - s + 1}$ $\displaystyle \implies \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac {m x^{m - 1} } {a s} \left({a x + b}\right)^{n - s + 1} + \frac {x^m} {a s} a \left({n - s + 1}\right) \left({a x + b}\right)^{n - s}$ above rules and Product Rule for Derivatives $\displaystyle$ $=$ $\displaystyle \frac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s} \left({m \left({a x + b}\right) + a \left({n - s + 1}\right) x}\right)$ extracting $\dfrac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s}$ as a factor $(2):\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s} \left({m b + a \left({m + n - s + 1}\right) x}\right)$ rearranging

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

Then $s = m + n + 1$.

Thus $(2)$ after rearrangement becomes:

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

Then:

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

and:

 $\displaystyle v \frac {\d u} {\d x}$ $=$ $\displaystyle \left({a x + b}\right)^s \frac {m b x^{m - 1} \left({a x + b}\right)^{n - s} } {\left({m + n + 1}\right) a}$ $\displaystyle$ $=$ $\displaystyle \frac {m b x^{m - 1} \left({a x + b}\right)^n} {\left({m + n + 1}\right) a}$

Thus by Integration by Parts:

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

$\blacksquare$

## Proof 2

$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$

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

 $\displaystyle \int x^m \paren {a x + b}^n \rd x$ $=$ $\displaystyle \frac {\paren {a x + b}^{n + 1} \paren {1 x + 0}^m} {\paren {m + n + 1} a} - \frac {m \paren {b 1 - a 0} } {\paren {m + n + 1} a} \int \paren {a x + b}^n \paren {1 x + 0}^{m - 1} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$

$\blacksquare$