Answer
$$\int x \tan ^{-1} x=\frac{\left(x^{2}+1\right) \tan ^{-1} x}{2}-\frac{x}{2}+C $$
Work Step by Step
Given
$$\int x \tan ^{-1} x$$
So by partition technique
Let $$u=\tan ^{-1} x \Rightarrow du= \frac{1}{1+x^2}dx$$
$$dv=x dx \Rightarrow v= \frac{x^2}{2} $$
So, we get
\begin{aligned} I&=uv- \int vdu\\
&=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x\\
&=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}} d x\\
& =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int 1-\frac{1}{1+x^{2}} d x\\
&=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+C\\
&=\frac{\left(x^{2}+1\right) \tan ^{-1} x}{2}-\frac{x}{2}+C \end{aligned}