Primitive of Power of Root of a x + b

Theorem

 * $\ds \int \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 2} } {a \paren {m + 2} } + C$

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

Then: