# Question #9592b

Mar 3, 2017

To find the indefinite integral, let $x = \sec t$,

so that ${x}^{2} - 1 = {\tan}^{2} x$ and $\mathrm{dx} = \sec t \tan t \mathrm{dt}$.

Upon substitution we will get

$\int \frac{\mathrm{dx}}{{x}^{2} - 1} ^ \left(\frac{3}{2}\right) = \int \frac{\sec t \tan t}{\tan} ^ 3 t \mathrm{dt}$

$= \int \sec t {\cot}^{2} t \mathrm{dt}$

$= \int \csc t \cot t \mathrm{dt} = - \csc t + C$

Since $x = \sec t$, we have $\cos t = \frac{1}{x}$, so

$\sin t = \frac{\sqrt{{x}^{2} - 1}}{x}$ and

$- \csc t + C = - \frac{1}{\sin} t = - \frac{x}{\sqrt{{x}^{2} - 1}} + C$

If the integrand should be $\left\mid {x}^{2} - 1 \right\mid$, Then for $a < 1$, use

$\left\mid {x}^{2} - 1 \right\mid = \left(1 - {x}^{2}\right)$.

In this case, substitute $x = \sin t$ and integrate to get

$\frac{x}{\sqrt{{x}^{2} - 1}} + C$