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

## Theorem

$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$

## Proof 1

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

Substituting $n + 1$ for $n$:

 $\displaystyle \int x^m \paren {a x + b}^{n + 1} \rd x$ $=$ $\displaystyle \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x$ $=$ $\displaystyle \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \int x^m \paren {a x + b}^{n + 1} \rd x$ rearrangement $\displaystyle \leadsto \ \$ $\displaystyle \int x^m \paren {a x + b}^n \rd x$ $=$ $\displaystyle \frac {m + n + 2} {\paren {n + 1} b} \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$ rearrangement $\displaystyle \leadsto \ \$ $\displaystyle \int x^m \paren {a x + b}^n \rd x$ $=$ $\displaystyle \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$ rearrangement

$\blacksquare$

## Proof 2

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

Setting $a := 1, b := 0, p x + q := a x + b$:

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

$\blacksquare$