# Primitive of Root of a x + b over Power of x/Formulation 2

## Theorem

$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 5}\right) a} {\left({2 m - 2}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x$

## Proof

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

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

 $\ds \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x$ $=$ $\ds \int x^{-m} \left({a x + b}\right)^{1/2} \ \mathrm d x$ $\ds$ $=$ $\ds \frac {x^{-m+1} \left({a x + b}\right)^{1/2 + 1} } {\left({-m + 1}\right) b} - \frac {\left({-m + \frac 1 2 + 2}\right) a} {\left({-m + 1}\right) b} \int x^{-m + 1} \left({a x + b}\right)^{1/2} \ \mathrm d x$ $\ds$ $=$ $\ds -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({m - \frac 5 2}\right) a} {\left({m - 1}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x$ simplifying $\ds$ $=$ $\ds -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 5}\right) a} {\left({2 m - 2}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x$ multiplying top and bottom by $2$

$\blacksquare$