∫(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
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.
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.
To solve this integral, you can start by simplifying the expression in the denominator. By multiplying the numerator and denominator by e^2x, you can rewrite the expression as follows:
∫ (e^-x) / (4 - e^-2x) dx
= ∫ (e^x * e^-x) / (4e^2x - 1) dx
= ∫ 1 / (4e^2x - 1) dx
Now, let's focus on the integral expression 1 / (4e^2x - 1). To solve this, we can try using a substitution method. Let u = 4e^2x - 1:
Differentiating both sides with respect to x, we get du = 8e^2x dx.
Rearranging, we have dx = (1/8e^2x) du.
Substituting the expressions for dx and u into the integral, we get:
∫ 1 / (4e^2x - 1) dx
= ∫ (1/8e^2x) du / u
= (1/8) ∫ (1/u) du
Integrating (1/u) du gives us ln|u| + C, where C is the constant of integration. Substituting back u = 4e^2x - 1, we obtain:
(1/8) ln|4e^2x - 1| + C
However, we need to make sure the domain of the function is correct. The natural logarithm function has a domain restriction, as it exists only for positive values. So, we need to ensure that the expression inside the logarithm is positive.
For ln|4e^2x - 1| to be positive, the argument 4e^2x - 1 must be greater than zero. Solving this inequality, we have:
4e^2x - 1 > 0
4e^2x > 1
e^2x > 1/4
2x > ln(1/4)
x > (1/2) ln(1/4)
x > -ln(2)/4
Therefore, the correct domain for the expression is x > -ln(2)/4.
So, the final solution should be:
(1/8) ln|4e^2x - 1| + C, for x > -ln(2)/4.