The value of $\int\limits_0^1\log_e (\frac{1}{x}-1) dx$ is:

0

The correct answer is Option (2) → 0

Evaluate the integral:

$\int_{0}^{1} \log\!\left(\frac{1}{x}-1\right)\,dx$

Simplify the expression:

$\frac{1}{x}-1=\frac{1-x}{x}$

$\log\!\left(\frac{1-x}{x}\right)=\log(1-x)-\log(x)$

Thus:

$\int_{0}^{1}\log(1-x)\,dx-\int_{0}^{1}\log(x)\,dx$

Use known results:

$\int_{0}^{1}\log x\,dx=-1$

$\int_{0}^{1}\log(1-x)\,dx=-1$

Therefore:

$-1 - (-1)=0$

The value of the integral is 0.