Question

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

Answer 1

$$\begin{array}{c}Given,\\ I=\int \frac{ln\left(\frac{x-1}{x+1}\right)}{x^2-1}dx\\ Integrationofln\left(\frac{x-1}{x+1}\right)=t,\\ differentiationbothsideis,\left(\therefore fromquestiontool=\frac{u}{v}=\frac{u^{{}^{\prime }}v-u{v}^{{}^{\prime }}}{v^2}\right)\\ \frac{1}{\frac{x-1}{x+1}}\left(\frac{1(x+1)-(x-1)}{{(x+1)}^2}\right)\\ \left(\frac{x+1}{x-1}\right)\left(\frac{1(x+1)-(x-1)}{{(x+1)}^2}\right)dx=dt\\ \frac{1}{x-1}\left(\frac{2}{(x+1)}\right)dx=dt\\ \frac{dx}{x^2-1}=\frac{1}{2}dt\\ Now,\\ I=\frac{1}{2}\int t\cdot dt\\ =\frac{1}{2}\left(\frac{t^2}{2}\right)+c\\ \therefore \frac{1}{4}{\left(ln\frac{(x-1)}{x+1}\right)}^2+c\\ sothatthecorectAnsweeris\frac{1}{4}{\left(ln\frac{(x-1)}{x+1}\right)}^2+c\end{array}$$