find the antiderivative of x^2e^(3x)

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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yabo2k wrote:find the antiderivative of x^2e^(3x)

use integration by parts

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

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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yabo2k wrote:i get ((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/9)+C

close...

look at the last bit...(2e^3x)/9

would it be e^(3x)/3?

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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yabo2k wrote:would it be e^(3x)/3?

no

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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yabo2k wrote:find the antiderivative of x^2e^(3x)

I'll do it step by step

udv=(x^2)(e^3x/3)-(e^3x/3)(2xdx)

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

Then do I need to find the antiderivative of the second part (2xe^3x)dx/3) because there are two fnc (2x/3) and (e^3x)

u= 2x/3 dv= (e^3x)dx

du= (2/3)dx v= (e^3x/3)

am I going in the right direction?

udv=(x^2)(e^3x/3)-(e^3x/3)(2xdx)

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

Then do I need to find the antiderivative of the second part (2xe^3x)dx/3) because there are two fnc (2x/3) and (e^3x)

u= 2x/3 dv= (e^3x)dx

du= (2/3)dx v= (e^3x/3)

am I going in the right direction?

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

yabo2k wrote:I'll do it step by step

udv=(x^2)(e^3x/3)-(e^3x/3)(2xdx)

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

Then do I need to find the antiderivative of the second part (2xe^3x)dx/3) because there are two fnc (2x/3) and (e^3x)

u= 2x/3 dv= (e^3x)dx

du= (2/3)dx v= (e^3x/3)

am I going in the right direction?

yes

for the antiderivative of the second part then I would get

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

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