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

[yabo2k]

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

[Martingale]

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

use integration by parts

[yabo2k]

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

[Martingale]

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

close...

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

[yabo2k]

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

[Martingale]

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

no

[Martingale]

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

Spoiler:

[yabo2k]

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?

[Martingale]

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

[yabo2k]

for the antiderivative of the second part then I would get

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