Answer
$$\frac{x^{2}}{2} \tanh ^{-1} x+\frac{1}{2} x-\frac{1}{2} \tanh ^{-1} x+C$$
Work Step by Step
Given $$\int x\tanh ^{-1} x d x$$
Let
\begin{align*}
u&=\tanh ^{-1} x\ \ \ \ \ \ \ \ \ \ \ \ dv=xdx\\
du&= \frac{1}{1-x^2} d x\ \ \ \ \ \ \ \ \ \ \ \ dv=\frac{1}{2}x^2\\
\end{align*}
Then
\begin{aligned}
\int x \tanh ^{-1} x d x=& \frac{x^{2}}{2} \tanh ^{-1} x \cdot-\frac{1}{2} \int \frac{x^{2}}{1-x^{2}} d x \\
&=\frac{x^{2}}{2} \tanh ^{-1} x+\frac{1}{2} \int \frac{-x^{2}}{1-x^{2}} d x \\
&=\frac{x^{2}}{2} \tanh ^{-1} x+\frac{1}{2} \int \frac{1-x^{2}-1}{1-x^{2}} d x \\
&=\frac{x^{2}}{2} \tanh ^{-1} x+\frac{1}{2} \int 1 \, d x-\frac{1}{2} \int \frac{1}{1-x^{2}} d x \\
&=\frac{x^{2}}{2} \tanh ^{-1} x+\frac{1}{2} x-\frac{1}{2} \tanh ^{-1} x+C
\end{aligned}
