Posted by Katie on Sunday, December 30, 2012 at 5:23pm.
Integrate using integration by parts
(integral) (5x) e^3x
u = 5x
du = dx
dv = e^3x
v = 3e^3x
I wonder if this is right so far.
= uv  (integral) v du
= (5x)(3e^3x)  (integral) (3e^3x)
=(5x)(3e^3x) + (integral) (3e^3x)
= (5x)(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
= (5x)(3e^3x) + (integral)  (1/3)e^3x
= (5x)(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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