Primitive of Reciprocal of x by Power of Root of a x + b

Theorem

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

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


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

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