I'm having a had time with this problem. Could you please help with step by step resolution?
e^-x / e^-1 + 3 dx
Thank you..
The "dx" at the end makes me think you are supposed to integrate
[e^-x / e^-1] + 3 .
But that would make it a calculus, not a precalculus, question.
It is not clear what the question is. In any case,
e^-x / e^-1 = e^(1-x)
The integral of
e^(1-x) + 3 dx is
-e^(1-x) + 3x + C
Thank you. You helped greatly.
Certainly! I'd be happy to help you with that problem.
To solve the integral, we can start by rewriting the expression in a more manageable form. Since e^-1 is a constant, we can take it out of the denominator:
e^-x / (e^-1 + 3) dx
Now, let's focus on the denominator. We'll simplify it by finding a common denominator. Multiply both the numerator and the denominator by e so that the denominator becomes e:
e^-x * e / (e^-1 * e + 3 * e) dx
Next, simplify the expression further:
e^(1 - x) / (e^0 + 3e) dx
e^0 is equal to 1, so we can rewrite the expression as:
e^(1 - x) / (1 + 3e) dx
Now, we can solve the integral using basic integral rules. In this case, we have a simple exponential function raised to a power, so we'll use the power rule for integration.
The power rule states that if we have ∫x^n dx, where n is any real number except -1, then the result is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In our case, n is 1 - x:
∫(e^(1 - x) / (1 + 3e)) dx
To integrate, we apply the power rule:
(1 / (1 + 3e)) * ∫e^(1 - x) dx
The integral of e^(1 - x) is simply e^(1 - x), so we can rewrite the expression as:
(1 / (1 + 3e)) * e^(1 - x) + C
Where C is the constant of integration.
That's it! The final result of the integral is (1 / (1 + 3e)) * e^(1 - x) + C.