Please answer 16th question Share with your friends Share 0 Lovina Kansal answered this Dear student Let I=∫exx2tan-1x+tan-1x+1x2+1dx=∫exx2tan-1x+extan-1x+exx2+1dxTake u=extan-1x⇒dx=1extan-1x+exx2+1du⇒du=x2extan-1x+extan-1x+exx2+1dxSo,I=∫du=u+C=extan-1x+C Regards 0 View Full Answer