∫ (x+arctanx)/(1+x^2) dx令arctanx=u,则x=tanu,dx=sec²udu,代入原式得：∫ (x+arctanx)/(1+x^2) dx=∫[(tanu+u)/(1+tan²u)]sec²udu=∫(tanu+u)du=∫tanudu+∫udu=-lncosu+(1/2)u²+C=-ln[1/√(...