Primitive of Root of a x + b over x/Proof 2

Theorem

 * $\displaystyle \int \frac {\sqrt{a x + b} } x \ \mathrm d x = 2 \sqrt{a x + b} + b \int \frac {\mathrm d x} {x \sqrt{a x + b} }$

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: