求在(-1,1)上的∫1/(1+x^2)^2dx

问题描述:

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...