The value of `int _(-1)^(1) log ((2-x)/(2+x)) d x ` is equal to

Options

A. `(1)/(2)`
B. 1
C. `-1`
D. 0

Answer

To solve the integral \( I = \int_{-1}^{1} \log\left(\frac{2-x}{2+x}\right) \, dx \), we will determine whether the integrand is an even or odd function.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_{-1}^{1} \log\left(\frac{2-x}{2+x}\right) \, dx \]

2. **Check for Odd/Even Function**:
To determine if \( f(x) = \log\left(\frac{2-x}{2+x}\right) \) is odd or even, we need to compute \( f(-x) \):
\[ f(-x) = \log\left(\frac{2 - (-x)}{2 + (-x)}\right) = \log\left(\frac{2+x}{2-x}\right) \]

3. **Relate \( f(-x) \) to \( f(x) \)**:
We can express \( f(-x) \) in terms of \( f(x) \):
\[ f(-x) = \log\left(\frac{2+x}{2-x}\right) = -\log\left(\frac{2-x}{2+x}\right) = -f(x) \]
This shows that \( f(x) \) is an odd function.

4. **Evaluate the Integral**:
Since \( f(x) \) is an odd function and we are integrating over a symmetric interval \([-1, 1]\), we can conclude:
\[ I = \int_{-1}^{1} f(x) \, dx = 0 \]

### Final Answer:
\[ I = 0 \]