integral 0 to 1 x(tan-1x)2dx Share with your friends Share 1 Tushar Kohli answered this ∫xtan-1xdx=x22tan-1x2-∫1+x2-1tan-1x1+x2dx=x22tan-1x2-∫tan-1x dx+∫tan-1x1+x2dx∫tan-1x dx=xtan-1x-∫x1+x2dx=xtan-1x-12ln1+x2x22tan-1x2+xtan-1x-12ln1+x2+tan-1x22Evaluating this from 0 to 1 we get that:π42+π4-12ln 2 -14 View Full Answer Abhishek Tiwari answered this pi/2-log2 1 Abhishek Tiwari answered this take 2 as second function and tan-1x as first functions and then solve byBy parts -1