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

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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

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

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


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

From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:

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


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

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

$\blacksquare$