How do you integrate $int x^2e^x dx$ using integration by parts?

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Answer 1 · 2016-01-27T01:53:25

$intx^2e^xdx = x^2e^x - 2xe^x + 2e^x + c$

$intudv = uv - intvdu$

So, if we pick $u = x^2$ and $dv = e^x$ so we'll have $du = 2x$ and $v = e^x$

$intx^2e^xdx = x^2e^x - int2xe^xdx$

Now we pick $u = x$ and $dv = e^x$ , so $v = e^x$ and $du = 1$

$intx^2e^xdx = x^2e^x - 2(xe^x - inte^xdx)$

This last integral is tabled, so

$intx^2e^xdx = x^2e^x - 2xe^x + 2e^x + c$

How do you integrate int x^2e^x dx int x^2e^x dx using integration by parts?