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

## Theorem

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

## Proof

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

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

 $\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^m}$ $=$ $\displaystyle \int x^{-1} \left({a x + b}\right)^{-m/2}$ $\displaystyle$ $=$ $\displaystyle \frac {-x^0 \left({a x + b}\right)^{-m/2 + 1} } {\left({- \frac m 2 + 1}\right) b} + \frac {-1 - \frac m 2 + 2} {\left({- \frac m 2 + 1}\right) b} \int x^{-1} \left({a x + b}\right)^{- m/2 + 1} \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\left({m - 2}\right) b \left({a x + b}\right)^{\left({m - 2}\right)/2} } - \frac 1 b \int \frac {\mathrm d x} {x \left({a x + b}\right)^{\left({m - 2}\right) / 2} }$ simplifying $\displaystyle$ $=$ $\displaystyle \frac 2 {\left({m - 2}\right) b \left({\sqrt{a x + b} }\right)^{m - 2} } + \frac 1 b \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^{m - 2} }$

$\blacksquare$