由分步积分得原式=2x(e^x-1)^(1/2)-2∫(e^x-1)^(1/2)dx令t=(e^x-1)^(1/2) 则dt=e^xdx/2t=(t^2+1)dx/2t dx=2tdt/(t^2+1)∴∫(e^x-1)^(1/2)dx=∫2t^2dt/(t^2+1)=∫[2-2/(1+t^2)]dt=2t-2arctant+C=2(e^x-1)^(1/2)-2arct...