Primitive of Root of a x + b over Power of x/Formulation 1

Theorem

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

Proof
Let:

From Integration by Parts:
 * $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

from which: