Local integrability of two functions Why $\log|x|$ is a locally integrable function and $1/|x|$ is not? Id know how their graphs look like but I don't know what is the exact difference causing local integrability of the first one.

First: there are two obstacles to global integrability here: behavior at $0$ and behavior at $\infty$. The function $1/|x|$ fails in both regards, that is
$$
\int_0^b 1/|x| \,dx = \infty
$$
for any $b$, and 
$$
\int_a^\infty 1/|x|\,dx = \infty
$$
for any $a$.
Since we're only concerned with local integrability, we only examine the first situation (indeed, $\log$ will fail global integrability). Further, since it is clear that $\int_a^b \,dx$ will be finite for any $a>0$, $b<\infty$ for either of these functions, we only have to look at the local integrability around $a = 0$.
Typically one evaluates improper integrals in the following way:
$$
\lim_{\epsilon \to 0} \int_\epsilon^b 1/|x|\,dx = \lim_{\epsilon \to 0}\log x|_\epsilon^b=  \log(b) - \lim_{\epsilon \to 0}\log\epsilon = \log b - (-\infty) = \infty
$$
so this failed to be integrable in a neighborhood of $0$.
On the other hand, for $\log$, 
$$
\lim_{\epsilon \to 0} \int_\epsilon^b \log x\,dx = \lim_{\epsilon \to 0}\left[x\log x - x\right]_\epsilon^b=  b\log b - b - \lim_{\epsilon \to 0}\left(\epsilon\log\epsilon - \epsilon\right) = b\log b - b - \lim_{\epsilon \to 0} \epsilon \log \epsilon$$
finally, you just need to prove (you can use L'Hopital's rule)
$$
\lim_{\epsilon \to 0} \epsilon \log \epsilon = 0
$$
to show that this integral converges.
