Posted by Katie on .
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,
dv = e^3x
v = (1/3)e^3x !!!!!! 
Calculus II 
Katie,
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,
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,
ok, thank you very much. Happy New Year

Calculus II 
jaime,
x^2

Calculus II 
jaime,
x^2