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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
