Evaluate : `int_(0)^(1)log ((1)/(x) -1) dx`

Answer

To evaluate the integral \( I = \int_{0}^{1} \log\left(\frac{1}{x} - 1\right) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
First, we rewrite the integrand:
\[ \log\left(\frac{1}{x} - 1\right) = \log\left(\frac{1 - x}{x}\right) \]
Thus, we can express the integral as:
\[ I = \int_{0}^{1} \log\left(\frac{1 - x}{x}\right) \, dx \]

### Step 2: Use the Property of Integrals
We can use the property of integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we have \( a = 1 \):
\[ I = \int_{0}^{1} \log\left(\frac{1 - (1 - x)}{1 - x}\right) \, dx = \int_{0}^{1} \log\left(\frac{x}{1 - x}\right) \, dx \]

### Step 3: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{1} \log\left(\frac{1 - x}{x}\right) \, dx \)
2. \( I = \int_{0}^{1} \log\left(\frac{x}{1 - x}\right) \, dx \)

Adding these two equations:
\[ 2I = \int_{0}^{1} \left( \log\left(\frac{1 - x}{x}\right) + \log\left(\frac{x}{1 - x}\right) \right) \, dx \]
Using the logarithmic property \( \log a + \log b = \log(ab) \):
\[ 2I = \int_{0}^{1} \log(1) \, dx \]

### Step 4: Evaluate the Integral
Since \( \log(1) = 0 \):
\[ 2I = \int_{0}^{1} 0 \, dx = 0 \]
Thus, we have:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \boxed{0} \]