Primitive of Power of a x + b over Power of p x + q/Formulation 3

Theorem

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

Proof
Let:

Then:

Thus: