接下来，我们计算∫(0，1)ln(1+r^2)rdr。应用分部积分法，令u=ln(1+r^2)，dv=rdr，则du=(2r/(1+r^2))dr，v=(1/2)r^2。根据分部积分公式，有。∫ln(1+r^2)rdr = (1/2)r^2ln(1+r^2) - ∫(1/2)r^2(2r/(1+r^2))dr。化简后得：∫ln(1+r^2)rdr = (1/2)r^2ln(1+r^2) - ∫r^2/(1+r^2)dr。进一步化简，∫r^2/(1+r^2)dr = ∫(1 - 1/(1+r^2))dr = r - arctan(r) + C。因此，∫ln(1+r^2)rdr = (1/2)r^2ln(1+r^2) - (1/2)r^2 + r - arctan(r) + C。