How do you integrate by parts: #int x e^3xdx#?

1 Answer
Truong-Son N.
Jun 9, 2015

This seems to be:

#int xe^(3x)dx#

#e^3(x)# wouldn't make sense (#e# is a constant, not a function), and #e^3x ne (ex)^3# (improper implications from #trig^n(x) = (trigx)^n#). #e^(x^3)# would be very advanced to integrate, and would not be remotely easy by integration by parts. #x^2e^3# would be way too simple.

Assuming so...

Let:
#u = x#
#du = 1dx#
#dv = e^(3x)dx#
#v = 1/3e^(3x)#

#= uv - intvdu#

#= x/3e^(3x) - 1/3inte^(3x)dx#

#= x/3e^(3x) - 1/3[1/3e^(3x)] + C]#

#= x/3e^(3x) - 1/9e^(3x) + C#

or

#= e^(3x)/9 (3x - 1) + C#

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