求在（-1,1）上的∫1/（1+x^2）^2dx

∫1/（1+x^2）^2dx
令x=tant,dx=sec^2tdt,x=-1,t=-π/4,x=1,t=π/4
∫[-1,1]1/（1+x^2）^2dx
=∫[-π/4,π/4]1/sec^2tdt
=∫[-π/4,π/4]cos^2tdt
=1/2∫[-π/4,π/4](1+cos2t)dt
=(1/2t+1/4sin2t)[-π/4,π/4]
=π/4＋1/2

∫1/（1+x^2）^2dx=x/2(1+x^2)+arctanx/2+C
所求定积分值为 1/2+π/4