求tanx的反函数的平方的积分

∫(arctanx)^2dx
=x(arctanx)^2-∫[x*2arctanx*(1/1+x^2)]dx
=x(arctanx)^2-∫[2x/(1+x^2)]arctanxdx
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^2)]arctanxdx
=∫2utanudu
=∫tanud(u^2)
=u^2tanu-∫u^2/(1+u^2)du
=u^2tanu-∫(u^2+1-1)/(u^2+1)du
=u^2tanu-∫[1-1/(1+u^2)]du
=u^2tanu-u+arctanu
=x(arctanx)^2-arctanx+tanx+C
即∫(arctanx)^2dx=x(arctanx)^2-x(arctanx)^2+arctanx-tanx+C