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.
 A: Let $p(x),q(x)$ be polynomials without common factors such that $\ln x=p(x)/q(x)$.
$\lim_{x\to 0^+}\ln x=-\infty$ means $q(x)$ has a root at $0$, so $q(x)=x q_1(x)$ for some polynomial $q_1(x)$.
$\lim_{x\to 0^+} x \ln x=0$ means $q_1(x)$ does not have a root at $x=0$ and that $p(x)$ has a root at $0$, so $x$ divides $p(x)$, contradicting the assumption of no common factors!
A: 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$$\lim_{x\to+\infty}\frac{e^{P(x)}}xe^{R(x)}=0\times1=0.$$
A: 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 $x^{r+1}$.
A: Yet another way to get at it:  if $\int \frac{dx}{x}$ were rational, we could write
$\displaystyle \int \dfrac{dx}{x} = \dfrac{p(x)}{q(x)}, \tag 1$
with $p(x), q(x) \in \Bbb R[x]$.
First question:  how do we know we can take $p(x), q(x) \in \Bbb R[x]$, and not $p(x), q(x) \in \Bbb C[x]$?  Well, if $p(x), q(x) \in \Bbb C[x]$, we could write
$\dfrac{p(x)}{q(x)} = \dfrac{\bar q(x) p(x)}{\bar q(x) q(x)}, \tag 2$
with 
$\bar q(x) q(x) \in \Bbb R[x]$.  We then see that
$\Im(\bar q(x) p(x)) = 0, \tag 3$
since $\int \frac{dx}{x}$ is real.  Thus
$\dfrac{p(x)}{q(x)} = \dfrac{\Re(\bar q(x) p(x))}{\bar q(x) q(x)}, \tag 4$
the quotient of two real polynomials; so we might as well assume that
$p(x), q(x) \in \Bbb R[x] \tag 5$
from the beginning. 
We can differentiate (1) and obtain
$\dfrac{1}{x} = \dfrac{p'(x)q(x) - p(x)q'(x)}{q^2(x)}, \tag 6$
which we may re-write as
$q^2(x) = x(p'(x)q(x) - p(x)q'(x)). \tag 7$
Now it is easy to see that
$\deg(p'(x)q(x) - p(x)q'(x)) = \deg p(x) + \deg q(x) - 1, \tag 8$
and so
$\deg x(p'(x)q(x) - p(x)q'(x)) = \deg p(x) + \deg q(x); \tag 9$
also,
$\deg q^2(x) = 2 \deg 2q(x), \tag{10}$
so we find
$2\deg q(x) = \deg p(x) + \deg q(x), \tag{11}$
whence
$\deg q(x) = \deg p(x). \tag{12}$
Now say
$p(x) =\sum_0^n p_i x^i, \tag{13}$
and
$q(x) = \sum_0^n q_i x^i; \tag{14}$
then
$\dfrac{p(x)}{q(x)} = \dfrac{\sum_0^n p_i x^i}{\sum_0^n q_i x^i} = \dfrac{x^n\sum_0^n p_i x^{i - n}}{x^n\sum_0^n q_i x^{i - n}} = \dfrac{\sum_0^n p_i x^{i - n}}{\sum_0^n q_i x^{i - n}}, \tag{15}$
whence
$\lim_{x \to \infty}\dfrac{p(x)}{q(x)} = \lim_{x \to \infty} \dfrac{\sum_0^n p_i x^{i - n}}{\sum_0^n q_i x^{i - n}} = \dfrac{p_n}{q_n} < \infty; \tag{16}$
but this contradicts
$\lim_{x \to \infty}\displaystyle \int \dfrac{dx}{x} = \lim_{x \to \infty} \ln x = \infty;  \tag{17}$
thus (1) is impossible.
A: If ln(x) is rational, consider what happens for large x.
Depending on the degrees of the numerator and denominator, it can go tp zero, tend tp a constant, or grow like an integral power of x.
But ln does not behave like any of these.
Therefore it can't be rational.
