Finding the definite integral $\int_0^1 \log x\,\mathrm dx$ $$\int_{0}^1 \log x \,\mathrm dx$$
How to solve this? I am having problems with the limit $0$ to $1$. Because $\log 0$ is undefined.

The only place there is a problem
is as $x \to 0$,
so let's look at
$\int_{c}^1 \log x \,\mathrm dx$
and see what happens as 
$c \to 0$.
Since $(x \ln x)' = 1+\ln x$,
$(x \ln x - x)' = \ln x$,
so
$\int_{c}^1 x \ln x dx = 
x \ln x - x |_c^1
= c-1-c \ln c
$.
As you have shown,
$\lim_{x \to 0} x \ln x
= \lim_{x \to 0} -x
= 0
$,
so
$\lim_{c \to 0}\int_{c}^1 x \ln x dx = 
x \ln x - x |_c^1
= \lim_{c \to 0}c-1-c \ln c
= -1
$.
