设√(e^x-1)=t,则dx=2tdt/(1+t²)
∵当x=ln2时,t=1.当x=0时,t=0
∴原式=2∫(0,1)t²dt/(1+t²)
=2∫(0,1)(1-1/(1+t²))dt
=2(t-arctant)|(0,1)
=2(1-π/4)
=2-π/2