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Let $I=\int x \tan ^{-1} x d x$

Taking $\tan ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$I=\tan ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int x d x\right\} d x$

$=\tan ^{-1} x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{1+x^{2}} \cdot \frac{x^{2}}{2} d x$

$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x$

$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int\left(\frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}}\right) d x$

$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int\left(1-\frac{1}{1+x^{2}}\right) d x$

$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{C}$

$=\frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+\mathrm{C}$