TITLE: Why isn't $\int \frac{1}{x}~dx = \frac{x^0}{0}$?
QUESTION [30 upvotes]: I know that $\int \frac{1}{x}~dx = \ln|x| + C$ and I know the antiderivative method works for all powers of $x$ except $-1$. But why is that the case? I am still in high school and teachers aren't really helpful with these questions. 
Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation.

REPLY [4 votes]: Notice that, because integration involves addition of a constant, the following is also an antiderivative of $x^n$:
$$\frac{x^{n+1}-1}{n+1}$$
It is indeterminate at $n=-1$, but one might instead consider
$$\lim_{n\to -1}\frac{x^{n+1}-1}{n+1}$$
which, by L'Hopital's rule, is equal to
$$\begin{align}
\lim_{n\to -1}\frac{x^{n+1}-1}{n+1}
&=\lim_{n\to -1}\frac{e^{(n+1)\ln(x)}-1}{n+1}\\
&=\lim_{n\to -1}\frac{e^{(n+1)\ln(x)}\ln(x)}{1}\\
&=e^0\ln(x)\\
&=\ln(x)\\
\end{align}$$