Derive the formula $$\bbox[5px, border:1px solid black, #F9F7ED]{ \int \tan ^{-1}x\,dx=x\,\tan ^{-1}x-\dfrac{1}{2}\ln \,(1+x^{2})+C }$$

Solution We use the integration by parts formula with $$u=\tan ^{-1}x\qquad \hbox{and }\qquad dv=dx$$

Then $$du=\frac{1}{1+x^{2}}\,dx\qquad \hbox{and}\qquad v=\int dx=x$$ and $$\int \tan ^{-1}x\,dx=x\cdot \tan ^{-1}x-\int \frac{x}{1+x^{2}}dx$$

To find the integral $\int \dfrac{x}{1+x^{2}}dx$, we use the substitution $t=1+x^{2}$. Then $dt=2x\,dx$, or equivalently, $x~dx=\dfrac{dt}{2}.$ $$\begin{equation*} \int \frac{x}{1+x^{2}}dx=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\ln \vert \,t\vert =\frac{1}{2}\ln (1+x^{2}) \end{equation*}$$

As a result, $\int \tan ^{-1} x \,dx=x \tan ^{-1}x-\dfrac{1}{2}\ln (1+x^{2})+C$.