∫ e^-x / 4-e^-2x dx
answer: ________+C
∫ 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
perhaps you mean
∫ e^-x / ( 4-e^-2x ) dx
or
∫ (( e^-x / 4) -e^-2x ) dx
I forgot the parantheses... ∫(e^-x)/ (4-e^(-2x)) dx
To find the integral of the given function, we can use the method of substitution. Let's break down the steps to solve it.
Step 1: Identify the appropriate substitution.
In this case, we can choose u = 4 - e^(-2x), as it simplifies the expression inside the integral.
Step 2: Find the derivative of u with respect to x.
Differentiating u = 4 - e^(-2x) gives us du/dx = 2e^(-2x).
Step 3: Rearrange the equation to solve for dx.
Rewriting the above equation, we have du = 2e^(-2x) dx. Dividing both sides by 2e^(-2x), we get dx = (1/2e^(-2x)) du.
Step 4: Substitute the variables in the integral.
Using the substitutions from steps 1 and 3, the integral becomes:
∫ e^(-x) / (4 - e^(-2x)) dx = ∫ (1/u) (1/2e^(-2x)) du = (1/2) ∫ du/u.
Step 5: Evaluate the simplified integral.
The integral of du/u is ln|u| + C, where C is the constant of integration.
Step 6: Substitute back the value of u.
Substituting back u = 4 - e^(-2x), the final answer is:
(1/2) ln|4 - e^(-2x)| + C.
Therefore, the answer to the given integral is (1/2) ln|4 - e^(-2x)| + C.