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

Theorem

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

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


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

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