∫[0，1]√x/(1+√x)dx
设，√x=t
则，∫√x/(1+√x)dx=∫t /(1+t)d(t^2)
=∫2t^2 /(1+t)dt
=∫[2t^2+2t-(2t+2)+2] /(1+t)dt
=∫(2t^2+2t)/(1+t) -(2t+2)/(1+t)+2/(1+t)dt
=∫2t -2+2/(1+t) dt
=t^2-2t+2ln|1+t|+c
=x-2√x+2ln|1+√x |+c
∴∫[0，1]√x/(1+√x)dx=1-2√1+2ln|1+√1 | - ( 0-2√0+2ln|1+√0 |)
=2ln2-1
