# 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 $s \in \Z$.

 $\ds v$ $=$ $\ds \paren {a x + b}^s$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \frac {\d v} {\d x}$ $=$ $\ds a s \paren {a x + b}^{s - 1}$ Power Rule for Derivatives and Derivatives of Function of $a x + b$

Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.

Then:

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

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

Then $s = m + n + 1$.

Thus $(2)$ after rearrangement becomes:

$\dfrac {\d u} {\d x} = \dfrac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}$

Then:

 $\ds u v$ $=$ $\ds \frac {x^m} {a s} \paren {a x + b}^{n - s + 1} \paren {a x + b}^s$ $\ds$ $=$ $\ds \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a}$

and:

 $\ds v \frac {\d u} {\d x}$ $=$ $\ds \paren {a x + b}^s \frac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}$ $\ds$ $=$ $\ds \frac {m b x^{m - 1} \paren {a x + b}^n} {\paren {m + n + 1} a}$

Thus by Integration by Parts:

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

$\blacksquare$

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

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

$\blacksquare$