Help with intergration?

I have to intergrate the following ae^-kx , 1/e^kx and 4/the square root of x. Can anyone give me a site with tutorials that can help me to do these?

Answers:
I like penguins
sorry, i was thinking it'd be social and political integration ,but i cant count above 15. sorry
1)int(a*e^(-kx)dx)
= -ae^(-kx)/k+C
2)int(e^(-kx)dx) = -e^(-kx)/k +B
3)int(4/sqrtx dx) =4*2sqrtx+A
=8sqrtx+A

A,B and C are constants and standard
integration tables were used

i hope that this helps
1) integral (ae^-kx) = (-a/k)e^(-kx) + A

2) integral(1/e^kx) = integral(e^(-kx)) = (-1/k)e^(-kx) + B

3) integral(4/the square root of x) = integral(4x^(-1/2)) = 4/(1/2)x^(1/2) = 8x^(1/2) + C = 8*(the square root of x) + C

for arbitrary constants A, B, C
The first two are essentially the same since 1/e^kx =e^(-kx). Also, write 4/sqrt(x) as 4*x^(-1/2).

In each case, you want to find a function whose derivative is the function you have. Since you know that the derivative of e^x is e^x, and since you know the chain rule, you can see that the derivative of e^(-kx) is -k*e^(-kx). Thus, the integral (not intergral) of ae^(-kx) will be (-a/k)*e^(-kx)+C. Also the integral of 4*x^(-1/2) will be 8*x^(1/2)=8sqrt(x)+C.

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