Primitive of Power of x over Root of a x + b

Theorem

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

Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:


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

Putting $n = -\dfrac 1 2$: