Question

$\int \frac{ln\left(\frac{x-1}{x+1}\right)}{x^2-1}dx$ is equal to

Answer 1

Let $\ln \left(\frac{x-1}{x+1}\right)=t\Rightarrow \frac{x+1}{x-1}\frac{((x+1)-(x-1))}{{(x+1)}^2}dx=dt\Rightarrow \frac{2}{x^2-1}dx=dt\Rightarrow \frac{dx}{x^2-1}=\frac{dt}{2}\Rightarrow \int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1}dx\Rightarrow \frac{\ln \left(t\right)}{2}dt\Rightarrow \frac{1}{2}(t\ln \left(t\right)-t)+C\Rightarrow \ln \left(\frac{x-1}{x+1}\right)\ln \ln \left(\frac{x-1}{x+1}\right)-\ln \left(\frac{x-1}{x+1}\right)+C$