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My attempt: Proof by contradiction: Restriction: for any function $h(x)$ inside $g(x)$, and for any function $j(x)$ inside $f(x)$, $j(x)$ or $f(x)$ may not be inverse functions of $h(x)$ or $g(x)$. $$f(x)^{g(x)}=e^{g(x)\ln(f(x))}\rightarrow\int e^{g(x)\ln(f(x))}dx=$$I then realized it was impossible to use the same method as is used to find the derivative, to prove that it's impossible to find the integral. I now tried to turn $f(x)^{g(x)}$ into a Taylor series, to somehow prove that it wasn't integrable. However as the derivative is:$$f(x)^{g(x)}\left(f'(x)\frac{g(x)}{f(x)}+g'(x)\ln(f(x))\right)$$ I quickly realized that this method, too, is unviable. Please Help!

Zuter_242
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    I don't know if Liouville's theory (see for example here) can answer your quest, but it is essential to know its existence. – Jean Marie May 10 '21 at 20:53
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    Liouville's theory is probably a good place to start. However note that if $f(x):=e^x$ and $g(x):=\ln(x)/x$, then $f(x)^{g(x)}=x$ is integrable. So you need some restrictions – Alex R. May 10 '21 at 21:00
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    I don't think you can expect to find a really simple proof. See here: https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – Hans Lundmark May 11 '21 at 06:44

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