令√x=t
x=t^2
x=0,t=0,x=1,t=1
dx=2tdt
∫[0,1]1/(1+√x)dx
=∫[0,1]2tdt/(1+t)
=2∫[0,1] [1-1/(1+t)]dt
=2[t-ln(1+t)][0,1]
=2-2ln2