質問: How do you integrate #int x^2e^(-3x)# by integration by parts method? =====================================================================
回答: #= -e^(-3x) ( 1/3 x^2 + 2/9 x + 2/27) + C# #### Explanation: #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#