Answer
$$\frac{1}{2}\left(\left(x^{2}+1\right) \tan ^{-1} x-x\right)+C$$
Work Step by Step
Use the formula
$$ \int x^{n} \tan ^{-1} a x d x=\frac{x^{n+1}}{n+1} \tan ^{-1} a x-\frac{a}{n+1} \int \frac{x^{n+1} d x}{1+a^{2} x^{2}}, \quad n \neq-1$$
with $n=a=1$
\begin{align*}
\int x \tan ^{-1} x d x&=\int x^{1} \tan ^{-1}(1 x) d x\\
&=\frac{x^{1+1}}{1+1} \tan ^{-1}(1 x)-\frac{1}{1+1} \int \frac{x^{1+1} d x}{1+(1)^{2} x^{2}}\\
&=\frac{x^{2}}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^{2} d x}{1+x^{2}}\\
&=\frac{x^{2}}{2} \tan ^{-1} x-\frac{1}{2} \int \left(1-\frac{ 1}{1+x^{2}}\right)dx
\end{align*}
Then use
$$\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C$$
with $a=1$, hence
\begin{align*}
\int x \tan ^{-1} x d x &=\frac{x^{2}}{2} \tan ^{-1} x-\frac{1}{2} \int \left(1-\frac{ 1}{1+x^{2}}\right)dx\\
&=\frac{x^{2}}{2} \tan ^{-1} x-\frac{1}{2} \int d x+\frac{1}{2} \int \frac{1}{1+x^{2}} d x\\
&=\frac{x^{2}}{2} \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+C\\
&=\frac{1}{2}\left(\left(x^{2}+1\right) \tan ^{-1} x-x\right)+C
\end{align*}
