令√(e^x-1)=t，则：x=0时，t=0；x=ln2时，t=1 且，===> e^x-1=t^2 ===> e^x=t^2+1 ===> x=ln(t^2+1) ===> dx=[1/(t^2+1)]*2tdt=[2t/(t^2+1)]dt 原定积分=∫t*[2t/(t^2+1)]dt =2∫[t^2/(t^2+1)]dt =2∫[(t^2+1)-1]/(t^2+1)dt =2∫[1-1/(t^2+1)]dt =2[t-arctant]| =2*[(1-π/4)-(0-0)] =2-(π/2).
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