∫ e^-x / 4-e^-2x dx

Question

answer: ________+C

Answer 1

∫ e^-x / (4 - e^-2x) dx
let u = e^-x, so du = -e^-x
now you have
∫ -1/(4-u^2) du = 1/4 ∫ (1/(2-u) + 1/(2+u)) du
= 1/4 ln|4-u^2|
= 1/4 ln|4-e^-2x| + C
or -1/2 tanh^-12x + C

Answer 2

perhaps you mean
∫ e^-x / ( 4-e^-2x ) dx
or
∫ (( e^-x / 4) -e^-2x ) dx

Answer 3

I forgot the parantheses... ∫(e^-x)/ (4-e^(-2x)) dx