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 $u$

$\blacksquare$


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