1.∫(e^x-1)^1/2dx 设（e^x-1)^1/2=t t^2=e^x-1,x=ln(t^2+1) ∫(e^x-1)^1/2dx =∫tdln(t^2+1) =∫2t^2/(t^2+1)dt =∫(2t^2+2-2)/(t^2+1)dt =2-2∫1/(t^2+1)dt+C =2-2arctan(t) +C =2-2arctan[（e^x-1)^1/2] +C 2.∫c...