How do you integrate $int x^2e^(2x)$ by parts?

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Answer 1 · 2017-01-24T20:44:31

$int x^2 e^(2x) dx = e^(2x)/4 (2x^2 -2x+ 1) +C$

Take $x^2$ as finite part, so in the integration by parts the degree of $x$ decreases:

$int x^2 e^(2x) dx = 1/2 int x^2 d(e^(2x)) = (x^2e^(2x))/2 - int xe^(2x) dx$

We can now solve the resulting integral by parts again:

$int xe^(2x) dx = 1/2 int x d(e^(2x)) = (xe^(2x))/2 - 1/2 int e^(2x) dx$

and we can now solve the last integral directly:

$int e^(2x) dx = 1/2e^(2x) + C$

Putting it all together:

$int x^2 e^(2x) dx = (x^2e^(2x))/2 - (xe^(2x))/2 + 1/4e^(2x) +C$

How do you integrate int x^2e^(2x)int x^2e^(2x) by parts?