Posted by **Katie** on Sunday, December 30, 2012 at 5:23pm.

Integrate using integration by parts

(integral) (5-x) e^3x

u = 5-x

du = -dx

dv = e^3x

v = 3e^3x

I wonder if this is right so far.

= uv - (integral) v du

= (5-x)(3e^3x) - (integral) (-3e^3x)

=(5-x)(3e^3x) + (integral) (3e^3x)

= (5-x)(3e^3x) + 9e^3x + C

- Calculus II -
**Damon**, Sunday, December 30, 2012 at 5:36pm
dv = e^3x

v = (1/3)e^3x !!!!!!

- Calculus II -
**Katie**, Sunday, December 30, 2012 at 6:34pm
Okay then, is this the final answer

= (5-x)(3e^3x) + (integral) -- (1/3)e^3x

= (5-x)(3e^3x) + (1/9)e^3x + C

- Calculus II -
**drwls**, Sunday, December 30, 2012 at 6:47pm
No. You are using the wrong v function in the u*v term. See Damon's answer.

The (Integral)-v*du term is:

-(Integral)(1/3)e^3x(-dx)

= (1/9)e^3x

So your answer is ok for that term

- Calculus II -
**Katie**, Sunday, December 30, 2012 at 7:02pm
ok, thank you very much. Happy New Year

- Calculus II -
**jaime**, Sunday, September 13, 2015 at 11:55pm
x^2

- Calculus II -
**jaime**, Sunday, September 13, 2015 at 11:57pm
x^2

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