\(\int \tan ^{-1} x \cdot d x\)
$$
=\int x \tan ^{-1} x \cdot \int \tan ^{-1} d x+c
$$
Apply Rule of integration by parts by taking \(\tan ^{-1} x\) as the first function.
$$
\begin{aligned}
&=\tan ^{-1} x x-\int \frac{1}{1+x^{2}} d x+c \\
&=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x+c \\
&=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^{2}\right|+c \\
&\int \tan ^{-1} x d x=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+c
\end{aligned}
$$
