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
-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!
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.
-(x-1)-1 or -1/(x-1)
The integral of esec(x) dx is not a function that may be expressed in terms of well-studied mathematical functions, elementary or nonelementary. In general, it must be evaluated by numerical methods.
integral of e to the power -x is -e to the power -x
if you mean e to the x power times log of x, it is e to the x divided by 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!
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.
∫ ex dx = ex + CC is the constant of integration.
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
-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!
x=1
(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.
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.
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.
You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.