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#