Asked by Monica
∫ e^-x / 4-e^-2x dx
answer: ________+C
answer: ________+C
Answers
Answered by
oobleck
∫ 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<sup><sup>-1</sup></sup>2x + C
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<sup><sup>-1</sup></sup>2x + C
Answered by
Anonymous
perhaps you mean
∫ e^-x / ( 4-e^-2x ) dx
or
∫ (( e^-x / 4) -e^-2x ) dx
∫ e^-x / ( 4-e^-2x ) dx
or
∫ (( e^-x / 4) -e^-2x ) dx
Answered by
Monica
I forgot the parantheses... ∫(e^-x)/ (4-e^(-2x)) dx
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