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
We start with the integral:
\[ I = \int_{0}^{1} \log\left(\frac{1}{x} - 1\right) \, dx \]
We can simplify the logarithm:
\[ \log\left(\frac{1}{x} - 1\right) = \log\left(\frac{1 - x}{x}\right) = \log(1 - x) - \log(x) \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{1} \left(\log(1 - x) - \log(x)\right) \, dx \]

### Step 2: Split the integral
Now, we can split the integral into two parts:
\[ I = \int_{0}^{1} \log(1 - x) \, dx - \int_{0}^{1} \log(x) \, dx \]

### Step 3: Evaluate the first integral
The integral \( \int_{0}^{1} \log(1 - x) \, dx \) can be evaluated using integration by parts. Let:
- \( u = \log(1 - x) \) and \( dv = dx \)
Then, \( du = -\frac{1}{1 - x} \, dx \) and \( v = x \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int_{0}^{1} \log(1 - x) \, dx = \left[ x \log(1 - x) \right]_{0}^{1} - \int_{0}^{1} \frac{x}{1 - x} \, dx \]
The boundary term evaluates to \( 0 \) since \( \log(1 - x) \) approaches \( 0 \) as \( x \) approaches \( 1 \) and is \( 0 \) at \( x = 0 \).

Thus, we need to evaluate:
\[ \int_{0}^{1} \frac{x}{1 - x} \, dx \]

### Step 4: Evaluate the second integral
The integral \( \int_{0}^{1} \log(x) \, dx \) can also be evaluated using integration by parts:
Let:
- \( u = \log(x) \) and \( dv = dx \)
Then, \( du = \frac{1}{x} \, dx \) and \( v = x \).

Using integration by parts:
\[ \int_{0}^{1} \log(x) \, dx = \left[ x \log(x) \right]_{0}^{1} - \int_{0}^{1} 1 \, dx \]
The boundary term evaluates to \( 0 \) (as \( x \log(x) \) approaches \( 0 \) as \( x \) approaches \( 0 \) and is \( 0 \) at \( x = 1 \)). Thus:
\[ \int_{0}^{1} \log(x) \, dx = 0 - 1 = -1 \]

### Step 5: Combine results
Now we combine our results:
\[ I = \int_{0}^{1} \log(1 - x) \, dx - (-1) \]
We know from properties of definite integrals that:
\[ \int_{0}^{1} \log(1 - x) \, dx = -1 \]
Thus:
\[ I = -1 + 1 = 0 \]

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