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

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Theorem

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


Proof 1

Let $s \in \Z$.

\(\ds v\) \(=\) \(\ds x^s\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d v} {\d x}\) \(=\) \(\ds s x^{s - 1}\) Power Rule for Derivatives


Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.

Then:

\(\ds u\) \(=\) \(\ds \frac {x^m \paren {a x + b}^n} {s x^{s - 1} }\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {x^{m - s + 1} \paren {a x + b}^n} s\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac {x^{m - s + 1} a n \paren {a x + b}^{n - 1} } s + \frac {\paren {m - s + 1} x^{m - s} \paren {a x + b}^n} s\) Primitive of Power of $a x + b$ and Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {a n x + \paren {m - s + 1} \paren {a x + b} }\) extracting $\dfrac {x^{m - s} \paren {a x + b}^n} s$ as a factor
\(\ds \) \(=\) \(\ds \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {a n x + a m x - a s x + a x + m b - s b + b}\) multiplying out
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {\paren {m + n + 1 - s} a x + \paren {m - s + 1} b}\) gathering terms


Let $s$ be selected such that $m + n + 1 - s = 0$.

Then $s = m + n + 1$.

Thus $(2)$ after rearrangement becomes:

$\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}$


Then:

\(\ds u v\) \(=\) \(\ds \frac {x^{m - s + 1} \paren {a x + b}^n} {m + n + 1} x^s\)
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1}\)

and:

\(\ds v \frac {\d u} {\d x}\) \(=\) \(\ds x^s \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}\)
\(\ds \) \(=\) \(\ds \dfrac {x^m} {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - \paren {m + n + 1} + 1} b}\)
\(\ds \) \(=\) \(\ds \dfrac {x^m} {m + n + 1} \paren {a x + b}^{n - 1} \paren {-n b}\)


Thus by Integration by Parts:

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

$\blacksquare$


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

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

$\blacksquare$


Sources

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