Question

不定积分1/（1+eˣ）dx？

Answer 1

∫1/(1+eˣ)dx
=∫eˣ/[eˣ(1+eˣ)]dx
=∫[(1+eˣ)-eˣ]/[eˣ(1+eˣ)]deˣ
=∫[(1/eˣ)-1/(1+eˣ)]deˣ
=x-ln[(1+eˣ)+C

Answer 2

∫ dx/(1+e^x)
=∫ e^(-x)/[e^(-x).(1+e^x)] dx
=∫ e^(-x)/[ 1+e^(-x)] dx
=-∫ d[1+e^(-x)]/[ 1+e^(-x)]
=-ln|1+e^(-x)| +C
=ln| 1/[1+e^(-x)]| +C
=ln| e^x/(1+e^x)| +C