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Hi there, use the substitution:
u = e^x
du = e^x * dx

Sorry I will be typing in short posts, because it is easier for me to see what I'm doing

\[1/ e ^{x} \int\limits_{}^{} e^x * dx/ ( 1 + e^x) = 1 / e^x \int\limits_{}^{} du / (1+u)\]

Now remember we let u = e^x, so plug that into the 1/e^x and move everything into the integral

\[\int\limits_{}^{}du / [u(1+u)]\]

Now use partial fractions:
( A / u ) + ( B / 1 + u ) = 1 / [u(1+u)]

x - ln (1+e^x) +C

you there?

yeah tryin to figure out ur question