Answer
$$\frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2}x+\frac{1}{4}\ln|1-x|-\frac{1}{4}\ln|1+x|+C $$
Work Step by Step
Given $$ \int x \tanh ^{-1} x d x$$Let
\begin{align*}
u&= \tanh ^{-1} x \ \ \ \ \ \ \ \ \ dv=x dx\\
du&= \frac{1}{1-x^2}dx \ \ \ \ \ \ \ \ \ dv=\frac{1}{2}x^{2}
\end{align*}
Then
\begin{align*}
\int x \tanh ^{-1} x d x &=
\frac{1}{2} x^{2} \tanh ^{-1} x-\frac{1}{2} \int \frac{x^{2}}{1-x^{2}} d x\\
&=\frac{1}{2} x^{2} \tanh ^{-1} x-\frac{1}{2} \int\left[-1+\frac{1}{1-x^{2}}\right] d x\\
&=\frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2} \int 1 \, d x-\frac{1}{2} \int \frac{1}{1-x^{2}} d x
\end{align*}
Since
\begin{aligned}
\frac{1}{1-x^{2}} &=\frac{1}{(1-x)(1+x)} \\
&=\frac{A}{(1-x)}+\frac{B}{(1+x)} \\
1 &=A(1+x)+B(1-x)
\end{aligned}
At $x=1$, $A=1/2$ and at $x= -1$, $B= 1/2$, so
\begin{align*}
\int \frac{1}{1-x^{2}} d x &=\frac{1}{2}\int \frac{dx}{(1-x)}+\frac{1}{2}\int \frac{dx}{(1+x)}\\
&=-\frac{1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C
\end{align*}
Hence
$$ \int x \tanh ^{-1} x d x = \frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2}x+\frac{1}{4}\ln|1-x|-\frac{1}{4}\ln|1+x|+C $$
