Primitive of Power of a x + b over Power of p x + q/Formulation 1

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Theorem

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


Proof

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

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


Setting $n := -n$:

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

$\blacksquare$


Sources