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Looking for help in solving a certain integral by parts:

$$ \int x^2 e^{-x} \, dx $$

What I did was:

$$ u = x^2 \text{ and } dv = e^{-x} \, dx$$

along with

$du=2x \, dx$ and $v = -e^{-x}$

The integral then would be $ \int x^2 e^{-x} \, dx = -x^2e^{-x} - \int -2xe^{-x} \, dx$

I simplified it to $-x^2e^{-x} - x^2e^{-x} + C$

My final answer was $-2^2e^{-x} +C$

and the book's answer is $-(x^2 + 2x + 2)e^{-x} + C$

I'm not entirely sure what I did wrong, hopefully somebody can point it out to me!

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  • $\begingroup$ How do you get your final answer? $\endgroup$ – user251257 Sep 29 '15 at 2:32
  • $\begingroup$ Question edited to show work - I just simplified it a bit $\endgroup$ – bankey Sep 29 '15 at 2:32
  • $\begingroup$ You need to integrate by parts a second time. Where you said "I simplified it to....", it's not at all clear how you got rid of the integral sign. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 29 '15 at 3:25
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You already have $$\int x^2e^{-x}dx = -x^2e^{-x} + 2\int xe^{-x}dx.$$

Now, you can continue $$\int xe^{-x}dx = -xe^{-x} + \int e^{-x}dx = -xe^{-x} - e^{-x}.$$

From this, you can get the final answer.

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  • $\begingroup$ So do I need to integrate by parts a second time? How do you go about figuring out when you need to do it a second or third time?? $\endgroup$ – bankey Sep 29 '15 at 2:35
  • $\begingroup$ Yes, you should. It depends what you want. $\endgroup$ – GAVD Sep 29 '15 at 2:41
  • $\begingroup$ Is there a way of knowing when you should continue integrating by parts? $\endgroup$ – bankey Sep 29 '15 at 2:42
  • $\begingroup$ It is a difficult question. Sometimes, you will fall in the infinitive loop of intergrating. :) $\endgroup$ – GAVD Sep 29 '15 at 2:44
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    $\begingroup$ Let $u = x$ and $v = -e^{-x}$, then you have $du = dx$ and $dv = e^{-x}dx$. From this, you can get that equation. $\endgroup$ – GAVD Sep 29 '15 at 2:49

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