$\int_0^1 \log \left(\frac{1}{x}-1\right) \mathrm{d} x=$

The integral $I = \int_0^1 \log \left(\frac{1}{x} - 1\right) \, \mathrm{d}x$ is evaluated by simplifying the logarithmic argument to $\log \left(\frac{1-x}{x}\right)$.
Applying the substitution property $\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x$ with $a=0$, $b=1$ yields $I = \int_0^1 \log \left(\frac{x}{1-x}\right) \, \mathrm{d}x$.

Adding both forms of $I$ gives $2I = \int_0^1 \left[ \log \left(\frac{1-x}{x}\right) + \log \left(\frac{x}{1-x}\right) \right] \, \mathrm{d}x = \int_0^1 \log(1) \, \mathrm{d}x$.

Since $\log(1) = 0$, it follows that $2I = 0$, and thus $I = 0$.

Final answer: $\boxed{0}$