I = ∫√xdx/(1+√x)
= ∫2xd√x/(1+√x)
= 2∫（x-1+1）d√x/(1+√x)
= 2∫（√x-1）d√x + 2∫d(1+√x)/(1+√x)
= 2[x/2-√x] + 2[ln(1+√x]
= -1+2ln2