∫arctanx\1+x^2dx

∫arctanx\1+x^2dx
∫arctanx\1+x^2dx

∫arctanx\1+x^2dx
显然d(arctanx)=1/(1+x²)dx
所以
∫arctanx /(1+x²)dx
=∫ arctanx d(arctanx)
=0.5(arctanx)²+C,C为常数