you really need to put an integral sign in therefor the antiderivative of the second part then I would get

(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)

- Martingale
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you really need to put an integral sign in therefor the antiderivative of the second part then I would get

(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)

I have them on my paper but I don't know how to put it into the computer. But I guess this is where I get stuck. I don't see how it can go any further than my first answer

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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I have them on my paper but I don't know how to put it into the computer. But I guess this is where I get stuck. I don't see how it can go any further than my first answer

you really need to put an integral sign in therefor the antiderivative of the second part then I would get

(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)

so did you find u, du, dv, and v again to get (2e^3x)/27

- Martingale
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no...not for the last part. just used the fact that one last timeso did you find u, du, dv, and v again to get (2e^3x)/27

so my final answer would be

((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C

((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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almost..check the sign on the (2e^3x)/27 termso my final answer would be

((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C

[/quote]

You have it as a subtraction. So where did i go wrong?

You have it as a subtraction. So where did i go wrong?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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You have it as a subtraction. So where did i go wrong?[/quote]

I have a subtraction for this small part of the problem. Combine this with the full problem.

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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did you see this post of mine?