# Primitive of Power of Root of a x + b

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## 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:

 $\displaystyle \int \left({\sqrt{a x + b} }\right)^m \ \mathrm d x$ $=$ $\displaystyle \frac 2 a \int u^{m + 1} \ \mathrm d x$ Primitive of Function of $a x + b$ $\displaystyle$ $=$ $\displaystyle \frac 2 a \frac {u^{m + 2} } {m + 2} + C$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \frac {2 \left({\sqrt{a x + b} }\right)^{m + 2} } {a \left({m + 2}\right)} + C$ substituting for $u$

$\blacksquare$