Question

∫(e^-x)/ (4-e^(-2x)) dx

the answer = 1/4 ln|4-e^-2x| + C seems to be wrong... I am not sure why though

Answer 1

it is correct, though you may want to try equivalent expressions, such as
1/4 (ln(2 - e^-x) + ln(2 + e^-x))
maybe with absolute-value signs; some say that
∫1/x dx = ln|x| rather than just lnx

Answer 2

disagree, taking the derivative of 1/4 ln|4-e^-2x| does not yield the original
expression. The original numerator would have had to contain e^(-2x)

I tried the following:
let u = e^-x , then du = -e^-x dx, or dx = du / -u

then ∫(e^-x)/ (4-e^(-2x)) dx ---> ∫ u/(4 - u^2) * du/-u
= ∫ -1/(4 - u^2) du

Using partial fractions:
let -1/(4 - u^2) = A/(2+u) + B/(2-u)
A(2-u) + B(2+u) = -1
let u = 2 ----> 4B = -1 , B = -1/4
let u = -2 ----> 4A = -1 , A = -1/4

then -1/(4 - u^2) = -1/4( 1/(2+u) + 1/(2-u)
1/(4 - u^2) = (1/4)( 1/(2+u) + 1/(2-u)

∫ -1/(4 - u^2) du = (1/4)( 1/(2+u) + 1/(2-u) )

= (1/4)( ln(2+u) <b>- ln(2 - u) )</b> + C
= (1/4)( ln(2 + e^-x) - ln(2 - e^-x) ) + C

looks like a minor error with a negative sign somewhere in Monica's solution,
I would have given her 9/10 for a similar solution.

Answer 3

thank you, I did make an error on the sign.
however when I enter the answer it says that the domain does not match the correct answer.