What is that different between $\int_0^1 \log(x)\, \mathrm dx$ and $ \int_0^1\frac{1}{x} \, \mathrm dx $?
While $\int_0^1\log(x) \, \mathrm dx =-1$, so it does converge, $\int_0^1\frac{1}{x}\, \mathrm dx $ diverges, but both $\log(x)$ and $\frac{1}{x}$ diverge at $x=0$.
I thought that both integrals are divergent because graph's area is infinite.
The surprising implication of these examples is that a Lebesgue integral $\int_a^b f(x) dx$ can converge even if either $f(a)$ or $f(b)$ diverges (or even both, if we replace $f(x)$ with $(f(x)+f(a+b-x))/2$). As you've noted, $\int_0^1\ln xdx=[x\ln x-x]_0^1=-1$ because $\lim_{x\to 0}x\ln x=\lim_{x\to 1}x\ln x=0$. Basically, the area is finite (or on a Riemann definition, undefined but certainly not $\pm\infty$), but that of a shape that becomes infinitely long at one end. That's not a contradiction. To take an example that's even easier, $\int_0^1x^{-1/2}dx=[2\sqrt{x}]_0^1=2$.