Integral[(dx/(ex+1)]
Here’s my method:
If
x = -Ln(u)
Then
dx = -du/u;
ex = 1/u;
u = e-x;
Substituting in the integral we get:
Integral[(-du/u)/((1/u)+1)] = -Integral[du/(1+u)]
Solving the integer in terms of u:
-Ln|1+u|+C
Substituting x in the result:
-Ln|1+e-x|+C
Sorry that i didn't use the...