Primitive of Power of Root of a x + b

Theorem

 * $\displaystyle \int \left({\sqrt{a x + b} }\right)^m \ \mathrm d x = \frac {2 \left({\sqrt{a x + b} }\right)^{m + 2} } {a \left({m + 2}\right)} + C$

Proof
Let $u = \sqrt{a x + b}$.

Then: