Find the integral?

Question

How to integrate e^(-2x)/[1+e^(-x)] dx?

 

Please show all your work. Thanks.

Answer 1

∫[e^-2x/(1+e^-x)dx = ∫[e^-x/(1+e^-x)]e^-xdx

 

    Let u = 1+e^-x.  Then e^-x = u-1 and du = -e^-xdx.

 

So, the integral can be rewritten as ∫[(u-1)/u](-du)

 

                                                = ∫[(1-u)/u]du

 

                                                = ∫[1/u - 1]du

 

                                                = ln lul - u + C

 

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

 

Since 1 + e^-x is always positive, we can remove the absolute value symbols.  Also, we can lump the constants together and relabel.

 

So, the simplified answer is ln(1+e^-x) - e^-x + C