这个定积分怎么算？看到标题了吗

原公式= ∫ (0，+∞)[1/(1+u)2]*[1/(1+u 2)]du

=∫(0，+∞)[(-u/2)/(1+u^2)+(1/2)/(1+u)+(1/2)/(1+u)^2]du

=[(-1/4)*ln(1+u^2)+(1/2)*ln(1+u)-(1/2)/(1+u)]|(0,+∞)

=(1/4)*{ln[(1+u)^2/(1+u^2)]-2/(1+u)}|(0,+∞)

=(1/4)*2

=1/2