答：
设t=√x,x=t^2；x=0,t=0；x=4,t=2
(0--4) ∫ [1/(1+√x)] dx
=(0--2) ∫ [1/(t+1)] d(t^2)
=(0--2) 2∫[(t+1-1)/(t+1) ] dt
=(0--2) 2t-2ln(t+1)
= 4-2ln3