How do you find the antiderivative of #x^2*e^(2x)#?

Question

1605 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T14:41:23

#= 1/2 e^(2x) (x^2 - x + 1/2 ) + C#

#int x^2 e^(2x) dx = int x^2 (1/2 e^(2x))^prime dx #

#= 1/2 x^2 e^(2x) - int (x^2)^prime * 1/2 e^(2x) dx #

#= 1/2 x^2 e^(2x) - int x e^(2x) dx #

#= 1/2 x^2 e^(2x) - int x (1/2 e^(2x))^prime dx #

#= 1/2 x^2 e^(2x) - ( 1/2 x e^(2x) - int (x)^prime * 1/2 e^(2x) dx )#

#= 1/2 x^2 e^(2x) - 1/2 x e^(2x) + int 1/2 e^(2x) dx #

#= 1/2 x^2 e^(2x) - 1/2 x e^(2x) + 1/4 e^(2x) + C#

#= 1/2 e^(2x) (x^2 - x + 1/2 ) + C#