Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x/Proof 2

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Theorem

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


Proof

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

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


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

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

$\blacksquare$