# ∫ 1 ( X 2 − 1 ) √ X 2 + 1 D X - Mathematics

Sum
$\int\frac{1}{\left( x^2 - 1 \right) \sqrt{x^2 + 1}} \text{ dx }$

#### Solution

$\text{ We have,}$
$I = \int \frac{dx}{\left( x^2 - 1 \right) \sqrt{x^2 + 1}}$
$\text{ Putting x }= \frac{1}{t}$
$\Rightarrow dx = - \frac{1}{t^2}dt$
$\therefore I = \int \frac{- \frac{1}{t^2}dt}{\left( \frac{1}{t^2} - 1 \right) \sqrt{\frac{1}{t^2} + 1}}$
$= \int \frac{- \frac{1}{t^2} dt}{\frac{\left( 1 - t^2 \right)}{t^2} \times \frac{\sqrt{1 + t^2}}{t}}$
$= \int \frac{- t \text{ dt }}{\left( 1 - t^2 \right) \sqrt{1 + t^2}}$
$\text{ Putting 1 }+ t^2 = u^2$
$\Rightarrow t^2 = u^2 - 1$
$\Rightarrow 2t\text{ dt }= 2u \text{ du }$
$\Rightarrow t \text{ dt } = u \text{ du }$
$I = - \int\frac{u \text{ du}}{\left( 1 - u^2 + 1 \right)u}$
$= - \int \frac{du}{2 - u^2}$
$= - \int \frac{du}{\left( \sqrt{2} \right)^2 - u^2}$
$= - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{u + \sqrt{2}}{u - \sqrt{2}} \right| + C$
$= - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{\sqrt{1 + t^2} + \sqrt{2}}{\sqrt{1 + t^2} - \sqrt{2}} \right| + C$
$= - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{\sqrt{1 + \frac{1}{x^2}} + \sqrt{2}}{\sqrt{1 + \frac{1}{x^2}} - \sqrt{2}} \right| + C$
$= - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{\sqrt{x^2 + 1} + \sqrt{2}x}{\sqrt{x^2 + 1} - \sqrt{2}x} \right| + C$

Concept: Integrals of Some Particular Functions
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 10 | Page 196