求不定积分∫dx/(根号下1+e^x)

Answer 1

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