(ex)3=e3x, so int[(ex)3dx]=int[e3xdx]=e3x/3
the integral ex^3 involves a complex function useful only to integrations such as this known as the exponential integral, or En(x). The integral is:-(1/3)x*E2/3(-x3). To solve this integral, and for more information on the exponential integral, go to http://integrals.wolfram.com/index.jsp?expr=e^(x^3)&random=false
(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.
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
The integral would be 10e(1/10)x+c
integral of e to the power -x is -e to the power -x
A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))
(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.
maths signs
The integral of cosine cubed is sinx- 1/3 sin cubed x + c
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
integral (a^x) dx = (a^x) / ln(a)
The integral would be 10e(1/10)x+c
integral of e to the power -x is -e to the power -x
better place to ask would be yahoo answers
replace square root o x with t.
A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))
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.
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