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.

Let $f(x)=p(x)/q(x)$ in lowest terms and consider $f'(x)$. If $q(x)$ does not have the factor $x$
then the denominator of $f'(x)$ doesn't have $x$ as a factor. But if
$q(x)$ has a factor of $x^r$, then the denominator of $f'(x)$ has a factor
of $x^{r+1}$.
