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Q. The value of $\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1} d x$ is equal to

Integrals

Solution:

$I=\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1} d x $
put $\ln \left(\frac{x-1}{x+1}\right)=t \Rightarrow \frac{2}{x^2-1} d x=d t$
$\Rightarrow I=\int t \frac{d t}{2}=\frac{t^2}{4}+C=\log ^2\left(\frac{x-1}{x+1}\right)+C$
$=\frac{1}{4} \log ^2\left(\frac{x+1}{x-1}\right)+C$