令[(1+e^x)^(1/2)]=t,得到1+e^x=t^2,x=ln(t^2-1)原式则变为∫td[ln(t^2-1)]=∫2t^2/(t^2-1)dt=∫[2+(1/(t-1))-1/(t+1)]dt=2t+ln(t-1)+ln(t+1)+c代会原式得到2(1+e^x)+x+ln(e^x+2)+c