Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x/Proof 1

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

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

$\blacksquare$