Question

$\int \frac{1}{x^2-1}ln\frac{x-1}{x+1}dx$.

Answer 1

$\int \frac{1}{x^2-1}\ln \frac{x-1}{x+1}dx$

Let $\ln \frac{x-1}{x+1}=t\frac{dt}{dx}=\frac{d\ln \frac{(x-1)}{(x+1)}}{d\left(\frac{x-1}{x+1}\right)}\frac{d\left(\frac{x-1}{x+1}\right)}{dx}=\frac{x+1}{x-1}\left[\frac{(x+1)\frac{d(x-1)}{dx}-(x-1)\frac{d(x+1)}{dx}}{{(x+1)}^2}\right]=\frac{[x+1-x+1]}{x^2-1}=\frac{2}{x^2-1}\frac{dt}{2}=\frac{dx}{x^2-1}=\int t\frac{dt}{2}=\frac{t^2}{2\times 2}=\frac{t^2}{4}=\frac{1}{4}{\left[\ln \frac{x-1}{x+1}\right]}^2+C$