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

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## Theorem

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

## Proof

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

Setting $n := -n$:

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

$\blacksquare$