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

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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

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

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


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

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

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


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

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

$\blacksquare$


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