int 1/(2 + e^x) dx

I can't figure out how to get the result (x - log(2 + e^x))/2 + C. It looks like integration by parts, it also looks similar to the integral of log(x), x(log(x) - 1) + C but I'm still completely lost. Should partial fractions be used?

Multiply top and bottom by e^(-x).

int 1/(2 + e^x) dx
That has got to be the worlds most difficult intelligence check ever.

>>3
So to do a check against an x of 6, you roll ~0.00247 6-sided dice, right?
Man, they don't call it Advanced Dungeons & Dragons for nothing.

>>3
i c wut u did thar

substitution. u = 2 + e^x, the integral becomes 1/(u^2 - 2u) du.