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How would I solve the following

$$\int x^3 e^{-x^2}\,dx$$

I set $u=e^{-x^2}(-2x)$ $du=e^{-x^2}(2x)$ $dv=x^3$ $v=\frac{x^4}{4}$

Then I did

$$e^{-x^2}\frac{x^4}{4}+\frac{1}{2}\int x^4e^{-x^2}(-x) \, dx$$

But I find myself stuck unfortunatly.

Michael Hardy
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Fernando Martinez
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3 Answers3

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Hint: Make the preliminary substitution $t=x^2$. You will end up with something familiar.

But if you wish you can let $u=x^2$ and $dv=xe^{-x^2}\,dx$. The integration by parts will proceed smoothly.

André Nicolas
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  • Yes I think I will take one try let me see. – Fernando Martinez Sep 06 '13 at 19:47
  • I think the final answer shall be $x^2-\frac{1}{2}e^{-x}+1(-1/2)e^{-x^2}+C$ – Fernando Martinez Sep 06 '13 at 19:49
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    I get $-\frac{1}{2}x^2e^{-x^2}-\frac{1}{2}e^{-x^2}+C$ – André Nicolas Sep 06 '13 at 19:56
  • +1. Maybe I could impose on you for advice as to how one could naturally come up with the $t = x^2$ substitution. Thanks. Regards, –  Sep 11 '13 at 23:22
  • Well, if one is looking for a "$u$-substitution," the general rule is that you let $u$ be whatever is Ugly. Anyway, $u=x^2$ is tempting. It would be nice if its derivative $2x$ were sitting in front. It sort of is, disguised as $x$, with another $x^2$ thrown in, but that's just $u$, so apart from a constant we will be looking at $ue^u$, doable, or at least simpler. We do all this while scanning the function, without writing anything. And then when it looks as if it will work, we start doing the details. – André Nicolas Sep 11 '13 at 23:30
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\begin{align} \int x^{3}{\rm e}^{-x^{2}}\,{\rm d}x &= \left.-\,{\partial \over \partial\mu} \int x{\rm e}^{-\mu x^{2}}\,{\rm d}x\right\vert_{\mu = 1} = \left.-\,{\partial \over \partial\mu} \left(-\,{{\rm e}^{-\mu x^{2}} \over 2\mu}\right)\right\vert_{\mu = 1} \\[3mm]&= \left.\left(% -\,{x^{2}{\rm e}^{-\mu x^{2}} \over 2\mu} - {{\rm e}^{-\mu x^{2}} \over 2\mu^{2}} \right)\right\vert_{\mu = 1} = -\,{x^{2}{\rm e}^{-x^{2}} \over 2} - {{\rm e}^{-x^{2}} \over 2} = \color{#ff0000}{\large% -\,{1 \over 2}\left(x^{2} + 1\right){\rm e}^{-x^{2}}} \end{align}

Felix Marin
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Hint: $$\int x^3 e^{-x^2}dx=\frac12 \int x^2e^{-x^2}d(x^2)=\frac12 \int ye^{-y}dy.$$

njguliyev
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