# Calculus II

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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 -

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

• Calculus II -

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 -

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

• Calculus II -

ok, thank you very much. Happy New Year

• Calculus II -

x^2

• Calculus II -

x^2

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