# How do you find the integral int1/(x^2*sqrt(x^2-9))dx ?

Aug 29, 2014

$= \frac{1}{9 x} \cdot \sqrt{{x}^{2} - 9} + c$, where $c$ is a constant

Explanation :

$= \int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 9}} \mathrm{dx}$

$= \int \frac{1}{{x}^{3} \sqrt{1 - \frac{9}{x} ^ 2}} \mathrm{dx}$

let's assume $\frac{3}{x} = t$, $- \frac{3}{x} ^ 2 \mathrm{dx} = \mathrm{dt}$

$= \int - \frac{t}{9 \sqrt{1 - {t}^{2}}} \mathrm{dt} = - \frac{1}{9} \int \frac{t}{\sqrt{1 - {t}^{2}}} \mathrm{dt}$

again assuming ${t}^{2} = z$, $2 t \mathrm{dt} = \mathrm{dz}$

$= - \frac{1}{9} \int \frac{1}{2} \cdot \frac{\mathrm{dz}}{\sqrt{1 - z}}$

$= - \frac{1}{18} \int \frac{1}{\sqrt{1 - z}} \mathrm{dz}$

$= - \frac{1}{18} \int {\left(1 - z\right)}^{- \frac{1}{2}} \mathrm{dz}$

$= \frac{1}{18} \cdot {\left(1 - z\right)}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + c$, where $c$ is a constant

Substituting value of $z$ and $t$ back,

$= \frac{1}{9} \sqrt{1 - \frac{9}{x} ^ 2} + c$, where $c$ is a constant

$= \frac{1}{9 x} \cdot \sqrt{{x}^{2} - 9} + c$, where $c$ is a constant