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

Theorem

 * $\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x = -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

Proof
Let:

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

from which: