What is the antiderivative of $(e^x)^2$

I know that the antiderivative of $e^x$ is just $e^x$... but what if it's $(e^x)^2$

I can see that the answer is $\frac{e^{2x}}{2}$... but how do you systematically find this? What is the rule?

I am going off of this video:

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My book doesn't even mention this in their charts:

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closed as too broad by Jwan622, Shailesh, JonMark Perry, Saad, GNUSupporter 8964民主女神 地下教會 Apr 18 '18 at 8:45

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    $\begingroup$ Your answer is wrong. Please look up the error function. en.wikipedia.org/wiki/Error_function $\endgroup$ – Yuriy S Apr 16 '18 at 14:05
  • $\begingroup$ You say that the antiderivative of $e^{x^2}$ is $\frac{e^{2x}}2$. Have you tried differentiating $\frac{e^{2x}}2$ and seen what you get? It's not $e^{x^2}$. @YuriyS Error function would be $e^{-x^2}$, while this is $e^{x^2}$. $\endgroup$ – Arthur Apr 16 '18 at 14:06
  • $\begingroup$ How did you come up with that answer? $e^{2x} \neq e^{x^2}$... $\endgroup$ – Andrew Li Apr 16 '18 at 14:06
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    $\begingroup$ That video has $(e^y)^2$ not $e^{y^2}$... $\endgroup$ – Andrew Li Apr 16 '18 at 14:14
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    $\begingroup$ @Jwan622 Not at all... the former is $e^y$ squared while the latter is $e$ to the power of $y^2$. $\endgroup$ – Andrew Li Apr 16 '18 at 14:16

You made a mistake assuming that $(e^{x})^2 = e^{x^2} $. They are widely different functions. $(e^{x})^2 \ne e^{x^2} $

$e^{x^2}$ does not have a simple primitive but $(e^{x})^2$ does.

let $I= \int (e^{x})^2\,dx$

$I = \int e^{2x}\,dx$

$I = \frac{e^{2x}}2+C$

You can verify this by finding the derivative of $\frac{e^{2x}}2$.

NOTE: $\int e^{x^2}\,dx$ is similar to the Gaussian integral. The indefinite integral does not have a value but the value of a definite integral of the same can be calculated using Polar coordinates


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