∫x^2/(x^2+1)dx
令x=tant dx=sect^2 t=arctanx
∫ x^2/(x^2+1)dx=∫tant^2/sect^4*sect^2dt
=∫tant^2/sect^2dt=∫sint^2dt
又因为cos2t=1-2sint^2 所以
sint^2=1-cos2t/2=1/2-1/2cos2t
∫sint^2dt=∫1/2-1/2cos2tdt
=1/2t-1/4sin2t+c 所以
∫x^2/(x^2+1)^2dx
=1/2arctanx-1/4sin(2arctanx)+c
∫
