integral of e to the power -x is -e to the power -x
The power law of indices says: (x^a)^b = x^(ab) = x^(ba) = (x^b)^a → e^(2x) = (e^x)² but e^x = 2 → e^(2x) = (e^x)² = 2² = 4
d/dx (e-x) = -e-x
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
e 2x = (1/2) e 2x + C ============
e^x/1-e^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!
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
e-2x^2 cannot be integrated, only approximated unless there is an additional x attached to the front of the e, otherwise this function is not integrable Actually, it can be integrated, but it requires multivariable calculus and a conversion from cartesian to polar form to do so.
int[e(2X) +e(- 2X)] integrate term by term 1/22 e(2X) - 1/22 e(- 2X) + C (1/4)e(2X) - (1/4)e(- 2X) + C ====================
integral of e to the power -x is -e to the power -x
x-1 = 1/x ∫1/x dx = ln x + 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.
The first derivative of e to the x power is e to the power of x.
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.
I can't integrate a-x /x-3 ?
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