Primitive of Reciprocal of Power of a x + b by Power of p x + q

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Theorem

$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$


Proof

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

$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$


Setting $m := -m$ and $n := -n$:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}\)
\(\displaystyle \) \(=\) \(\displaystyle \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\left({-n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{-m+1} \left({p x + q}\right)^{-n+1} - a \left({-m - n + 2}\right) \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n+1} \ \mathrm d x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)\)

$\blacksquare$


Sources