How do you integrate $int x^2e^(-3x)$ by integration by parts method?

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Answer 1 · 2016-07-31T16:29:30

$= -e^(-3x) ( 1/3 x^2 + 2/9 x + 2/27) + C$

$int x^2e^(-3x) \ dx$

$= int x^2 ( - 1/3 e^(-3x))^prime \ dx$

which by IBP, ie: $int u v' = uv - int u' v$

$= - 1/3 x^2 e^(-3x) - int ( x^2)^prime ( - 1/3 e^(-3x)) \ dx$

$= - 1/3 x^2 e^(-3x) + 1/3 int 2x e^(-3x) \ dx$

$= - 1/3 x^2 e^(-3x) + 2/3 int x ( - 1/3 e^(-3x))^prime \ dx$

which by IBP again
$= - 1/3 x^2 e^(-3x) + 2/3 ( - 1/3 x e^(-3x) - int ( x)^prime ( - 1/3 e^(-3x)) \ dx)$

$= - 1/3 x^2 e^(-3x) -2/9 x e^(-3x) + 2/9 int e^(-3x) \ dx$

$= - 1/3 x^2 e^(-3x) -2/9 x e^(-3x) + 2/9 ( - 1/3 e^(-3x)) + C$

$= - 1/3 x^2 e^(-3x) -2/9 x e^(-3x) - 2/27 e^(-3x) + C$

$= -e^(-3x) ( 1/3 x^2 + 2/9 x + 2/27) + C$

How do you integrate int x^2e^(-3x)int x^2e^(-3x) by integration by parts method?