Primitive of Power of a x + b over Power of p x + q/Formulation 2

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Theorem

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


Proof

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

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


Setting $n := -n$:

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

$\blacksquare$


Sources

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