The value of $\int_0^1 \log_e\left(\frac{1}{x} - 1\right)dx$ is:

The integral is $\int_0^1 \log_e\left(\frac{1}{x} - 1\right)dx$

Simplifying the expression inside the logarithm:

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

The integral becomes:

$\int_0^1 \ln\left(\frac{1-x}{x}\right)dx$

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Using the logarithm property $\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)$:

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

The integral splits into:

$\int_0^1 \ln(1-x)dx - \int_0^1 \ln(x)dx$

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For $\int_0^1 \ln(1-x)dx$, let $u = 1-x$

Then $du = -dx$

When $x = 0$: $u = 1$

When $x = 1$: $u = 0$

Substituting:

$\int_0^1 \ln(1-x)dx = \int_1^0 \ln(u)(-du)$

$= -\int_1^0 \ln(u)du$

$= \int_0^1 \ln(u)du$

$= \int_0^1 \ln(x)dx$

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Substituting back into the split integral:

$\int_0^1 \ln(1-x)dx - \int_0^1 \ln(x)dx = \int_0^1 \ln(x)dx - \int_0^1 \ln(x)dx$

$= 0$

Therefore, $\int_0^1 \log_e\left(\frac{1}{x} - 1\right)dx = 0$

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Alternate Solution:

Let the given integral be $I$:

$I = \int_{0}^{1} \log_e \left(\frac{1-x}{x}\right) dx$ --- (Equation 1)

Apply the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ by replacing $x$ with $(1 - x)$:

$I = \int_{0}^{1} \log_e \left(\frac{1-(1-x)}{1-x}\right) dx$

$I = \int_{0}^{1} \log_e \left(\frac{x}{1-x}\right) dx$ --- (Equation 2)

Add Equation 1 and Equation 2:

$2I = \int_{0}^{1} \left[ \log_e \left(\frac{1-x}{x}\right) + \log_e \left(\frac{x}{1-x}\right) \right] dx$

Using the logarithmic property $\log m + \log n = \log(m \cdot n)$:

$2I = \int_{0}^{1} \log_e \left(\frac{1-x}{x} \cdot \frac{x}{1-x}\right) dx$

$2I = \int_{0}^{1} \log_e(1) \, dx$

Since $\log_e(1) = 0$:

$2I = \int_{0}^{1} 0 \, dx \implies I = 0$