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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

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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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
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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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