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