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

Theorem

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

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


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

Putting $n := \dfrac m 2$ and $m := -1$: