答：
设√(1+e^x)=t>1,1+e^x=t^2,x=ln(t^2-1)
x=ln3,t=2
x=ln8,t=3
原式
=(2→3) ∫td[ln(t^2-1)]
=(2→3) ∫ [t*2t/(t^2-1)]dt
=(2→3) 2∫ [(t^2-1+1)/(t^2-1)]dt
=(2→3) 2∫ [1+1/(t^2-1)]dt
=(2→3) 2*[t+(1/2)ln|(x-1)/(x+1)|]
=2*[3+(1/2)ln(1/2)]-2*[2+(1/2)ln(1/3)]
=6-ln2-4+ln3
=2+ln3-ln2
=2+ln(3/2)