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I'm not sure if you mean e^x + 17 or e^(x+17) so we'll do both.
First, the integral of e^x + 17
because these terms are being added you can integrate them separately:
integral((e^x)dx) + integral(17dx)
integral of e^x is just e^x + C
Integral of 17 is 17x + C, so we get:
e^x + 17x + C
Second, the integral of e^(x+17)
we know how to integrate the form e^u, so just do a u substitution
u=x+17
du=dx
so we get
integral((e^u)du)=e^u + C
resubstitute for u and get e^(x+17) + C
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Integral of e to the power of x?
The antiderivative, or indefinite integral, of ex, is ex + C.
What is the antiderivative of x to the -1?
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
What is the antiderivative of exp -x?
It is -exp (-x) + C.
What is the antiderivative of -cscxcotx?
First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
7 plus 5k equals 8k plus 1 what is k?
k = 2 7 + (5 x 2) = 17 (8 x 2) + 1 = 17
What is the antiderivative of e to the -x?
-e-x + C.
What is the antiderivative of xex?
One can use integration by parts to solve this. The answer is (x-1)e^x.
What is the antiderivative of e to the power of one divided by -x?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
Integral of e to the power of x?
The antiderivative, or indefinite integral, of ex, is ex + C.
What is the antiderivative of e raised to the one tenth x?
The integral would be 10e(1/10)x+c
What is the antiderivative of x to the -1?
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
What is the antiderivative of e to the power of one divided by x?
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
What is the antiderivative of radical 1 plus 3 radical x?
sqrt(1) + 3*sqrt(x) = 1 + 3*x^1/2So the antiderivative is x + [3*x^(3/2)]/(3/2) + c = x + 2*x^(3/2) + c where c is the constant of integration.
How do you solve g x equals -3x plus 1?
If: x = -3x+1 Then: x+3x = 1 => 4x =1 So: x = 1/4 or 0.25 ----------- I notice that the question requests a solution for g x = -3x + 1. It seems possible that parentheses around the 'x' after the 'g' have gone missing, along with a prime indicating the derivative of the function g. This being the case, we would be seeking the antiderivative of -3x + 1. The antiderivative of a sum is the sum of the antiderivatives. So we can look at -3x and +1 separately. The derivative of x2 is 2x. Therefore, the antiderivative of x is x2/2, and the antiderivative of -3x is -3x2/2. The antiderivative of 1 is x. Overall, the solution is the antiderivative -3x2/2 + x + C, where C is an arbitrary constant.
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
What is the antiderivative of x to the negative 6 5ths?
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
