Prove the Integral of $\frac{1}{x}$ is not a Rational Function The title is pretty self explanatory, I'd like to prove that
\begin{align*}
\int\frac{1}{x}\,dx
\end{align*}
cannot be a rational function. I have attempted a proof by contradiction, but it doesn't seem to lead anywhere. If it is assumed that $F(x)=\frac{p(x)}{q(x)}$,  where $p(x)$ and $q(x)$ are polynomials, then using logarithmic differentiation,
\begin{align*}
\frac{1}{x}=\frac{p(x)}{q(x)}\left(\frac{p'(x)}{p(x)}-\frac{q'(x)}{q(x)}\right)
\end{align*}
I don't see how this leads to a contradiction. I get similar results using the quotient rule
\begin{align*}
\frac{1}{x}=\frac{p'(x)q(x)-p(x)q'(x)}{[q(x)]^2}
\end{align*}
Writing out the terms of $p(x)$ and $q(x)$ seems too messy. Any help or suggestions would be greatly appreciated.

As you wrote in a comment, this is the same thing as proving that $\log$ is not a rational function. Suppose it was. Then we could express $\log$ as $\frac pq$, where $p$ and $q$ are polynomial functions. Furthermore, $\deg p>\deg q$, since $\lim_{x\to+\infty}\log(x)=+\infty$. So $\frac pq=P+R$, where $P$ is a non constant polynomial function and $R$ is a rational function such that $\lim_{x\to+\infty}R(x)=0$. Besides,$$1=\frac{e^{\log x}}x=\frac{e^{P(x)}}xe^{R(x)}.$$This is impossible, since$$\lim_{x\to+\infty}\frac{e^{P(x)}}xe^{R(x)}=(+\infty)\times1=+\infty$$or$$\lim_{x\to+\infty}\frac{e^{P(x)}}xe^{R(x)}=0\times1=0.$$
