设 √(1+e^x) = u
1+e^x = u²
e^x dx = 2udu
dx = 2udu/e^x = 2udu/(u²-1)
∫dx/(1+e^x)
=∫2udu/[u(u²-1)]
=∫2du/[(u²-1)]
=∫[1/(u-1) - 1/(u+1)]du
=ln|u-1|-ln|u+1|+C
=ln|(u-1)/(u+1)|+C
=ln|[√(1+e^x) -1]/[√(1+e^x) +1]|+C
急,求原函数：1/（（1+e的x次方）的开根号）
设 √(1+e^x) = u
1+e^x = u²
e^x dx = 2udu
dx = 2udu/e^x = 2udu/(u²-1)
∫dx/(1+e^x)
=∫2udu/[u(u²-1)]
=∫2du/[(u²-1)]
=∫[1/(u-1) - 1/(u+1)]du
=ln|u-1|-ln|u+1|+C
=ln|(u-1)/(u+1)|+C
=ln|[√(1+e^x) -1]/[√(1+e^x) +1]|+C