Chapter 8: Techniques of Integration - Section 8.2 - Integration by Parts - Exercises 8.2 - Page 455: 11

$$\int \tan ^{-1} y \ dy=y\tan ^{-1} y - \frac{1}{2} \ln |1+y^{2}|+C $$

Given $$\int \tan ^{-1} y \ dy $$ Now, using integration by parts: $$ u=\tan ^{-1} y \Rightarrow du= \frac{1}{1+y^2}dy $$ $$ dv= dy \Rightarrow v= y $$ So, we get \begin{aligned} I&=\int \tan ^{-1} y \ dy\\ &=uv- \int vdu\\ &=y\tan ^{-1}y - \int \frac{y }{1+y^{2}} d x\\ &=y\tan ^{-1} y - \frac{1}{2}\int \frac{2y }{1+y^{2}} d x\\ &=y\tan ^{-1} y - \frac{1}{2} \ln |1+y^{2}|+C \end{aligned}