Question

用第一换元积分法求不定积分1/ex+1

如图第十四题

Answer 1

let
u=e^x
du = e^x dx
dx = du/u
∫dx/(e^x+1)

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