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

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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 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

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 $p := 1, q := 0, n := m, m := n$:

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

$\blacksquare$


Sources

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