Mathematics

# $\int { \sqrt { \left( \dfrac { x-1 }{ x+1 } \right) } } dx$

##### SOLUTION
$\int{(\sqrt{\dfrac{x-1}{x+1}})}dx$

This can be written as

$\Rightarrow \int{\sqrt{\dfrac{(x-1)(x-1)}{(x+1)(x-1)}}}dx$

$\rightarrow \int{\sqrt{\dfrac{(x-1)^2}{x^2-1}}}dx$

$\Rightarrow \int{{\dfrac{x-1}{\sqrt{x^2-1}}}}dx$

$\Rightarrow \int{\dfrac{x}{\sqrt{x^2-1}}}dx-\int{\dfrac{1}{\sqrt{x^2-1}}}dx$

In the first interval take $x^2-1=t\space\Rightarrow 2xdx=dt$

Second interval is in the form  of $\int\dfrac{1}{\sqrt{x^2-a^2}}dx=\log|\sqrt{x^2-a^2}+x|+C$

Using these results we get

$\Rightarrow \dfrac{1}{2}\int{\dfrac{dt}{\sqrt t}}-\log|\sqrt{x^2-1}+x|+C$

$\Rightarrow (\sqrt t)-\log|\sqrt{x^2-1}+x|+C$

$\Rightarrow (\sqrt {x^2-1})-\log|\sqrt{x^2-1}+x|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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On the basis of above information, answer the following questions :

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