The function `F(x)= int_(0)^(pi) "log" ((1-x))/((1+x)) dx` is a function which is

Options

A. even
B. odd
C. periodic
D. None of these

Answer

To solve the problem, we need to analyze the function defined by the integral:

\[ F(x) = \int_{0}^{\pi} \log\left(\frac{1 - x}{1 + x}\right) \, dx \]

We want to determine whether this function is even, odd, or periodic.

### Step-by-Step Solution:

1. **Define the Function**:
We start with the definition of the function:
\[ F(x) = \int_{0}^{\pi} \log\left(\frac{1 - x}{1 + x}\right) \, dx \]

2. **Substitute \( -x \)**:
To check if \( F(x) \) is odd, we need to compute \( F(-x) \):
\[ F(-x) = \int_{0}^{\pi} \log\left(\frac{1 - (-x)}{1 + (-x)}\right) \, dx = \int_{0}^{\pi} \log\left(\frac{1 + x}{1 - x}\right) \, dx \]

3. **Use Logarithmic Properties**:
We can rewrite the logarithm:
\[ F(-x) = \int_{0}^{\pi} \log\left(\frac{1 + x}{1 - x}\right) \, dx = \int_{0}^{\pi} -\log\left(\frac{1 - x}{1 + x}\right) \, dx \]
This simplifies to:
\[ F(-x) = -\int_{0}^{\pi} \log\left(\frac{1 - x}{1 + x}\right) \, dx = -F(x) \]

4. **Conclusion**:
Since \( F(-x) = -F(x) \), we conclude that \( F(x) \) is an odd function.

### Final Result:
Thus, the function \( F(x) \) is an odd function.