Primitive of Root of a x + b over x/Proof 1

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