∫dx/(1-e^x)^2不定积分求解!

Question

∫dx/(1-e^x)^2 不定积分求解!

Answer 1

∫dx/(1-e^x)^2
=∫dx/[1-2e^x+e^(2x)]
=∫e^(-x)*dx/[e^(-x)-2+e^x]
=-∫e^(-x)*d(-x)/[e^(-x)-2+e^x]
=-∫d[e^(-x)]/[e^(-x)-2+e^x]
设t=e^(-x)
原式=-∫dt/(t-2+1/t)
=-∫tdt/(t^2-2t+1)
=-∫(t-1+1)dt/(t-1)^2
=-∫dt/(t-1)-∫dt/(t-1)^2
=-ln|t-1|+1/(t-1)+C
=-ln|e^(-x)-1|+1/[e^(-x)-1]+C