∫(e^-x)/ (4-e^(-2x)) dx
The integral is solved as = (1/4)( ln(2 + e^-x) - ln(2 - e^-x) ) + C
but when I submit it it says that the domain does not match the correct answer...
I am not sure what to do
3 answers
i do gjhwre,kjgfhekjrhfnkjhsfnlkrwjdnvjqrhbdvnjrehnfjlemrfjnli3kr,ewc
well, the integration is certainly right, and since the domain is the same (x ≠ -ln2) I don't see the problem.
Now, if you write it as 1/2 tanh(2e^x), the domain is x < -ln2, then maybe that comes into play.
Now, if you write it as 1/2 tanh(2e^x), the domain is x < -ln2, then maybe that comes into play.
Monica, I believe both oobleck and I gave you an answer for this one when it was posted before.
But with Jiskha's current lack of a "Search" feature, I cannot find it.
I recall sending my answer through the "Wolfram" app and it was correct.
But with Jiskha's current lack of a "Search" feature, I cannot find it.
I recall sending my answer through the "Wolfram" app and it was correct.
1 answer