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
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