dv = e^3x
v = (1/3)e^3x !!!!!!
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
6 answers
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
= (5-x)(3e^3x) + (integral) -- (1/3)e^3x
= (5-x)(3e^3x) + (1/9)e^3x + C
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
The (Integral)-v*du term is:
-(Integral)(1/3)e^3x(-dx)
= (1/9)e^3x
So your answer is ok for that term
ok, thank you very much. Happy New Year
x^2
x^2