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,

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

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

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

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

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