So integrate (e^-3x)/(12 + e^-3x)?

What does our natural log rule say? u'/u = ln|u| + c, right? Like the integral of 1/x is ln|x|. You've got the same thing here. Only difference is, u' would equal -3e^-3x. Well, we have -1/3 of that. Fix this by throwing a -1/3 outside the integral sign.

Then it's just . Or -1/3ln|u| + c