# Primitive of Power of Root of a x + b over x squared

Jump to navigation Jump to search

## Theorem

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

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

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

$\blacksquare$