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I tried $u$-substiution with $u = x^2$ but that leaves me with $x$ values and not a simpler function to integrate.Is there a better way to integrate?

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    $\begingroup$ It can be proved that the antiderivative $x \, \mapsto \, \exp(x^2)$ cannot be expressed in terms of "elementary" functions. $\endgroup$
    – pitchounet
    Oct 14, 2014 at 22:12
  • $\begingroup$ Is your integration indefinite? Does it have any bound in integration? $\endgroup$
    – CLAUDE
    Oct 14, 2014 at 22:17
  • $\begingroup$ @AMIR No, but more specifically I'm trying to integrate $$x^5(e^(-x^2)$$ and integral $$3x^3(2^(x^2)$$ so I thought maybe integrating $$e^(x^2)$$ uses the same concepts. $\endgroup$
    – user159778
    Oct 14, 2014 at 22:18
  • $\begingroup$ put your mathematical expression in $$ in your comment to be readable $\endgroup$
    – CLAUDE
    Oct 14, 2014 at 22:20
  • $\begingroup$ @Ben You should open a new question stating the full problem. Someone will probably tell you to let $u=-x^2$ so that $x^4 x dx = -u^2\frac{1}{2}du$, or some such. This substitution reduces it to an integration by parts problem. $\endgroup$ Oct 14, 2014 at 22:22

2 Answers 2

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Represent $e^{x^2}$ with the Maclauren series, then separate the term $\frac{1}{k!}$. You should now have an equation that can be easily integrated using the power rule. Integrate it and multiply the $\frac{1}{k!}$ term back in. We could stop here, but if you notice it equates to something close to the error function, then you can say that it is equal to the error function times other missing terms.

$$e^{x^2}=\sum_{k=0}^\infty\frac{(x^2)^k}{k!}\\ \int e^{x^2}=\sum_{k=0}^\infty\frac{1}{k!}\int x^{2k}dx\\ \int e^{x^2}=\sum_{k=0}^\infty\frac{x^{2k+1}}{k!(2k+1)}\\ \sum_{k=0}^\infty\frac{x^{2k+1}}{k!(2k+1)}=\int_0^x e^{-t^2}dt\\ erf(x)=\frac{2}{\sqrt\pi}\int_0^x e^{t^2}dt\\ \int e^{x^2}=-\frac{i\sqrt\pi}{2}erf(ix)$$

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The result of integration can't be expressed using elementary functions. Such integrals can be calculated only on a defined segment and sometimes only numerically. In other words, there's no such function expressed as a trigonometrical, degree, exponential, logarithmic function, their sum, difference, multiplication, division or composition which derivative equals $e^{x^2}$.

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