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The antiderivative, or indefinite integral, of ex, is ex + C.

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What is the antiderivative of e to the x plus 17?

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


What is the integral of e to the negative x?

-e^(-x) or negative e to the negative x this is because you multiply the function (e) by: 1 / (the derivative of the power ... in this case: -1) e^(-x) * (1/-1) = -e^(-x) Don't forget to add your constant!


Integrate e raised to power x raised to power 2?

This integral cannot be performed analytically. Ony when the integral is taken from 0 to infinity can it be computed by squaring the integral and applying a change of variable (switching to polar coordinates). if desired I could show how to do this.


What is the integral of dx divided by x-1 to the power of 2?

-(x-1)-1 or -1/(x-1)


Integral of 1 divided by sinx cosx?

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

Related Questions

Integral of e to the power of -x?

integral of e to the power -x is -e to the power -x


What is the integral to negative e to the power x?

The integral of (-e^x) with respect to (x) is (-e^x + C), where (C) is the constant of integration. This represents the family of functions whose derivative is (-e^x).


What is an integral of ex logx?

if you mean e to the x power times log of x, it is e to the x divided by x


How do you integrate xe power x?

Use integration by parts. integral of xe^xdx =xe^x-integral of e^xdx. This is xe^x-e^x +C. Check by differentiating. We get x(e^x)+e^x(1)-e^x, which equals xe^x. That's it!


Integration of e e?

Writing equations in questions is problematic - some symbols regularly get eliminated.The integral of e to the power x is: e to the power x + C If your expression contains no variables, for example e times e, or e to the power e, then the entire expression is a constant; in this case, the integral is this constant times x + C.


What is the integral of e to the power of x with respect to x?

∫ ex dx = ex + CC is the constant of integration.


What is the antiderivative of e to the x plus 17?

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


What is the integral of e to the negative x?

-e^(-x) or negative e to the negative x this is because you multiply the function (e) by: 1 / (the derivative of the power ... in this case: -1) e^(-x) * (1/-1) = -e^(-x) Don't forget to add your constant!


What is the integral of the e?

x=1


What is the integral of e raised to x raised to 8?

(e^x)^8 can be written as e^(8*x), so the integral of e^(8*x) = (e^(8*x))/8 or e8x/ 8, then of course you have to add a constant, C.


What would be the integral of x where x has power 2?

If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.


Integrate e raised to power x raised to power 2?

This integral cannot be performed analytically. Ony when the integral is taken from 0 to infinity can it be computed by squaring the integral and applying a change of variable (switching to polar coordinates). if desired I could show how to do this.