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 integrand
We start by rewriting the integrand:
\[ I = \int_{0}^{1} \log\left(\frac{1}{x} - 1\right) \, dx = \int_{0}^{1} \log\left(\frac{1 - x}{x}\right) \, dx \]
This is achieved by simplifying \( \frac{1}{x} - 1 = \frac{1 - x}{x} \).

### Step 2: Use properties of logarithms
Using the property of logarithms, we can separate the terms:
\[ I = \int_{0}^{1} \left( \log(1 - x) - \log(x) \right) \, dx \]
This can be split into two separate integrals:
\[ I = \int_{0}^{1} \log(1 - x) \, dx - \int_{0}^{1} \log(x) \, dx \]

### Step 3: Evaluate the integrals
Let’s denote:
\[ I_1 = \int_{0}^{1} \log(1 - x) \, dx \quad \text{and} \quad I_2 = \int_{0}^{1} \log(x) \, dx \]

#### Evaluating \( I_2 \):
The integral \( I_2 \) can be evaluated using integration by parts:
Let \( u = \log(x) \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \).
\[ I_2 = \left[ x \log(x) \right]_{0}^{1} - \int_{0}^{1} x \cdot \frac{1}{x} \, dx \]
Evaluating the boundary term:
\[ \left[ x \log(x) \right]_{0}^{1} = 1 \cdot \log(1) - \lim_{x \to 0} x \log(x) = 0 - 0 = 0 \]
Thus,
\[ I_2 = 0 - \int_{0}^{1} 1 \, dx = -1 \]

#### Evaluating \( I_1 \):
The integral \( I_1 \) can be evaluated using the substitution \( x = 1 - t \):
\[ I_1 = \int_{0}^{1} \log(t) \, dt = -1 \quad \text{(by symmetry, similar to \( I_2 \))} \]

### Step 4: Combine the results
Now we can combine the results:
\[ I = I_1 - I_2 = (-1) - (-1) = -1 + 1 = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{0}^{1} \log\left(\frac{1}{x} - 1\right) \, dx = 0 \]