Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 37

$$\frac{-1}{2}[\tanh ^{-1} x]^2+C$$

Given $$\int \frac{\tanh ^{-1} x}{x^{2}-1} d x$$ Let $$u=\tanh^{-1}x\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ du=\frac{dx}{1-x^2} $$ Then \begin{align*} \int \frac{\tanh ^{-1} x}{x^{2}-1} d x&=\int -udu\\ &=\frac{-1}{2}u^2+C\\ &= \frac{-1}{2}[\tanh ^{-1} x]^2+C \end{align*}