Answer
\[I=x\tan^{-1}x-\frac{1}{2}\ln |1+x^2|+C\]where $C$ is constant of integration
Work Step by Step
Let \[I=\int \tan^{-1}x\:dx\]
\[I=\int \tan^{-1}x.(1)\:dx\]
Using integration by parts
\[I=\tan^{-1}x\int dx-\int \left((\tan^{-1}x)'\int dx\right)dx\]
\[I=x\tan^{-1}x-\int\frac{x}{1+x^2}dx\]
\[I=x\tan^{-1}x-\frac{1}{2}\int\frac{2x}{1+x^2}dx\]
Let $u=1+x^2\;\; $ then $\;du=2xdx$ ___(1)
\[I=x\tan^{-1}x-\frac{1}{2}\int\frac{1}{u}du\]
\[I=x\tan^{-1}x-\frac{1}{2}\ln |u|+C\]
Where $C$ is constant of integration
Using (1)
\[I=x\tan^{-1}x-\frac{1}{2}\ln |1+x^2|+C\]
Hence,
\[I=x\tan^{-1}x-\frac{1}{2}\ln |1+x^2|+C\]
