Integrate #xe^(4x)# ?

1 Answer
Eddie
Apr 7, 2017

#= 1/4 x e^(4x) - 1/16 e^(4x) + C#

Explanation:

#int x e^(4x) \ dx#

We will use IBP:

#= int \ x \ d/dx(1/4 e^(4x)) \ dx#

#= 1/4 x e^(4x) - 1/4int \ x' \ e^(4x) \ dx#

#= 1/4 x e^(4x) - 1/4int e^(4x) \ dx#

#= 1/4 x e^(4x) - 1/16 e^(4x) + C#

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