用分部积分法∫(arctanx)^2dx=x(arctanx)^2-∫[x*2arctanx*(1/1+x^2)]dx=x(arctanx)^2-∫[2x/(1+x^2)]arctanxdx再对∫[2x/(1+x^2)]arctanxdx先换元,再分部积分：令u=arctanx,du=1/(1+x^2)x=tanu,2x=2tanu∫[2x/(1+x^...