# Primitive of Power of x by Root of a x + b

## Theorem

$\displaystyle \int x^m \sqrt{a x + b} \ \mathrm d x = \frac {2 x^m} {\left({2 m + 3}\right) a} \left({\sqrt{a x + b} }\right)^3 - \frac {2mb} {\left({2 m + 3}\right) a} \int x^{m - 1} \sqrt{a x + b} \ \mathrm d x$

## Proof

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

Putting $n := \dfrac 1 2$:

 $\ds \int x^m \sqrt{a x + b} \ \mathrm d x$ $=$ $\ds \int x^m \left({a x + b}\right)^{1/2} \ \mathrm d x$ $\ds$ $=$ $\ds \frac {x^m \left({a x + b}\right)^{3/2} } {\left({m + \frac 1 2 + 1}\right) a} - \frac {m b} {\left({m + \frac 1 2 + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^{1/2} \ \mathrm d x$ $\ds$ $=$ $\ds \frac {x^m \left({\sqrt{a x + b} }\right)^3} {\left({m + \frac 3 2}\right) a} - \frac {m b} {\left({m + \frac 3 2}\right) a} \int x^{m - 1} \sqrt{a x + b} \ \mathrm d x$ simplifying $\ds$ $=$ $\ds \frac {2 x^m} {\left({2 m + 3}\right) a} \left({\sqrt{a x + b} }\right)^3 - \frac {2mb} {\left({2 m + 3}\right) a} \int x^{m - 1} \sqrt{a x + b} \ \mathrm d x$ multiplying top and bottom by $2$

$\blacksquare$