设√x=u,x=u^2,dx=2udu
∫(0,1)ln(1+√x)dx
=∫(0,1)2uln(1+u)du
=[u^2ln(1+u)](0,1)-∫(0,1)u^2/(1+u)du
=ln2-∫(0,1)[u-1+1/(u+1)]du
=ln2-[1/2u^2-u+ln(u+1)](0,1)
=ln2-(1/2-1+ln2)
=1/2
定积分（1,0）ln（1+根号x）dx
设√x=u,x=u^2,dx=2udu
∫(0,1)ln(1+√x)dx
=∫(0,1)2uln(1+u)du
=[u^2ln(1+u)](0,1)-∫(0,1)u^2/(1+u)du
=ln2-∫(0,1)[u-1+1/(u+1)]du
=ln2-[1/2u^2-u+ln(u+1)](0,1)
=ln2-(1/2-1+ln2)
=1/2