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Someone can help me with this integral please?

$$ \int ne^{-n^2} dn $$

I tried with parts

$$ u = e^{-n^2} \ \ \ \ \ \ \ \ \ \ v = n \ dn $$

$$ du = -e^{-n^2}2n \ \ \ \ \ \ \ \ \ \ \ v = \int n \ dn, \ \ \ v = \frac{n^2}{2} $$

Then,

$$ \int ne^{-n^2} dn = \frac{e^{-n^2}n^2}{2} + \int n^3e^{-n^2} dn $$

After, I tried with parts again, but I can't get to answer

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What is the derivative of $n \mapsto e^{-n^2}$? –  copper.hat Jul 3 at 2:35

2 Answers 2

Hint: Let $u=n^{2}$.${}{}{}{}{}{}{}{}{}$

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That should give the answer since then $du=2ndn$ so $\int ne^{-n^{2}}dn=\frac{1}{2}\int e^{-n^{2}}2ndn=\frac{1}{2}\int e^{-t}dt$. This last integral can be solved. –  user71352 Jul 3 at 2:38
    
You right, Thanks! –  user17629 Jul 3 at 2:40
    
You're welcome. –  user71352 Jul 3 at 2:40

Do $u = -n^2$, and you get $du = -2n\,dn$, and so

$$ \int\,ne^{-n^2}dn = -\dfrac{1}{2}\int\,e^u\,du$$

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