How do you integrate #int xe^(-3x)# by integration by parts method?

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Answer 1 · 2016-07-23T03:01:20

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

#int xe^(-3x) \ dx #

#= int x d/dx ( - 1/3e^(-3x) ) \ dx #

which, by IBP,

#= -1/3 x e^(-3x) - int d/dx( x) - 1/3e^(-3x) \ dx #

#= -1/3 x e^(-3x) +1/3 int e^(-3x) \ dx #

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

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