设√(1+e^x) = t,可知t>=1 则x = ln(t²-1) dx = 2tdt/(t²-1) ∫dx/√(1+e^x) =∫2tdt/t(t²-1) =∫2dt/(t²-1) =∫[1/(t-1) - 1/(t+1)]dt =ln(t-1) - ln(t+1) + C =ln[√(1+e^x) - 1] - ln[√(1+e^x) + 1] + C
