求∫1/(1+e^x)^2的不定积分

Question

求∫1/(1+e^x)^2的不定积分

Answer 1

∫1/(1+e^x)² dx
=∫e^x/[e^x(1+e^x)²] dx
=∫1/[e^x(1+e^x)²] de^x
令e^x=u
=∫1/[u(1+u)²] du
=∫(1/u-1/(1+u)-1/(1+u)²] du
=ln|u|-ln|u+1|+1/(1+u)+C
=x-ln|e^x+1|+1/(e^x+1)+C