Show that

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

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

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

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

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

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

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