Answer
$$x \tanh ^{-1} x+\frac{1}{2} \ln \left|1-x^{2}\right|+C$$
Work Step by Step
Given $$\int \tanh ^{-1} x d x$$
Let
\begin{align*}
u&=\tanh ^{-1} x\ \ \ \ \ \ \ \ \ \ \ \ dv=dx\\
du&= \frac{1}{1-x^2} d x\ \ \ \ \ \ \ \ \ \ \ \ dv=x\\
\end{align*}
then
\begin{aligned}
\int \tanh ^{-1} x d x &=x \tanh ^{-1} x-\int \frac{x}{1-x^2} d x\\
&=x \tanh ^{-1} x+\frac{1}{2} \ln \left|1-x^{2}\right|+C
\end{aligned}
