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

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Theorem

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


Proof 1

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

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


Substituting $m + 1$ for $m$:

\(\ds \int x^{m + 1} \paren {a x + b}^n \rd x\) \(=\) \(\ds \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \frac {\paren {m + 1} b} {\paren {m + n + 2} a} \int x^m \paren {a x + b}^n \rd x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\paren {m + 1} b} {\paren {m + n + 2} a} \int x^m \paren {a x + b}^n \rd x\) \(=\) \(\ds \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \int x^{m + 1} \paren {a x + b}^n \rd x\) rearranging
\(\ds \leadsto \ \ \) \(\ds \int x^m \paren {a x + b}^n \rd x\) \(=\) \(\ds \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x\) rearranging
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x\) rearranging

$\blacksquare$


Proof 2

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

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


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

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

$\blacksquare$