# Primitive of Root of a x + b over Power of x/Formulation 1

## Theorem

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

## Proof

Let:

 $\displaystyle u$ $=$ $\displaystyle \sqrt{a x + b}$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u}{\mathrm d x}$ $=$ $\displaystyle \frac a {2 \sqrt{a x + b} }$ Power Rule for Derivatives etc. $\displaystyle v$ $=$ $\displaystyle \frac {-1} {\left({m - 1}\right) x^{m - 1} }$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d v}{\mathrm d x}$ $=$ $\displaystyle \frac 1 {x^m}$ Power Rule for Derivatives

From Integration by Parts:

$\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

from which:

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

$\blacksquare$