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?

The so-called "deep reason" is not deep at all.  The term $\displaystyle \log(x)$ is simply the constant term in the expansion of $\displaystyle \frac{x^{n+1}}{n+1}$ around $n=-1$.  To see this, we simply write
$$\begin{align}\frac{x^{n+1}}{n+1}&=\frac{e^{(n+1)\log(x)}}{n+1}\\\\&=\frac{1}{n+1}\sum_{k=0}^\infty \frac{(n+1)^k\,\log^k(x)}{k!}\\\\&=\sum_{k=0}^\infty\frac{(n+1)^{k-1}\log^k(x)}{k!}\\\\&=\frac{1}{n+1}+\log(x)+\frac12(n+1)\log^2(x)+O((n+1)^2)\end{align}$$
whence we see the leading terms in the asymptotic ($n\sim -1$) expansion of $\displaystyle \frac{x^{n+1}}{n+1}$.  Obviously, we see that $\lim_{n\to -1}\frac{x^{n+1}-1}{n+1}=\log(x)$.

More simply, let $f(x,n)$ be given by the integral
$$\begin{align}
f(x,n)&=\int_1^x t^n\,dt\\\\
&=\frac{x^{n+1}-1}{n+1}\tag 1
\end{align}$$
Note that $f(x,n)$ is continuos on $(0,\infty)\times \mathbb{R}$ (i.e., $n$ need not be restricted to the integers).
Then, note that the limit as $n\to -1$ of $f(x,n)=f(x,-1)$ is
$$\begin{align}
\int_1^x \frac1t\,dt&=f(x,-1)\\\\
&=\lim_{n\to -1}f(x,n)\\\\
&=\lim_{n\to -1} \frac{x^{n+1}-1}{n+1}\\\\
&=\lim_{n\to -1}\frac{e^{(n+1)\log(x)}-1}{n+1}\\\\
&=\log(x)
\end{align}$$
So, we can recover the expected result, $\int_1^x \frac{1}{t}\,dt=\log(x)$, by taking the limit in $(1)$ for $n\ne -1$.
