The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$) In general, the indefinite integral of $x^n$  has power $n+1$.  This is the standard power rule.  Why does it "break" for $n=-1$?  In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ fails to hold for $n=0$.
Is there some deep reason for this discontinuity?

$\int x^n = \frac {x^{n+1}}{n+1} +c$
clearly is undefined if $n = -1$ so the rule is breaking down.
but since $x^{-1}$ is bounded and continuous if $x>0$ the integral should exist.
Lets tackle this from the other direction.
What is $\frac {d}{dx} \log x$?
First, $e = \lim_\limits{n\to\infty}(1+\frac 1n)^n$
From the definition of derivative.
$\frac {d}{dx} \log x =$$ \lim_\limits {h\to 0} \frac {\log (x+h) - \log x}{h}\\
\lim_\limits {h\to 0} \frac {\log (\frac {x+h}{x})}{h} \\
\lim_\limits {h\to 0}\frac {\log (1+\frac hx)}h\\
\lim_\limits {h\to 0} \frac 1x \frac xh \log (1+\frac hx)\\
\frac 1x\lim_\limits {h\to 0}  \log (1+\frac hx)^\frac xh\\
\frac 1x \log (\lim_\limits {h\to 0} (1+\frac hx)^\frac xh)$
We can say $n = \frac xh$ and and as $h$ goes to $0, n$ goes to $\infty$
$\frac {d}{dx} \log x = \frac 1x\log e\\
\frac {d}{dx} \ln x = \frac 1x$
