Question

换元积分法

Answer 1

如图

Answer 2

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