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 can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx \]

### Step 1: Substitute \( x \) with \( -x \)

We first compute \( I \) by substituting \( x \) with \( -x \):

\[ I = \int_{-1}^{1} \log\left(\frac{2 - (-x)}{2 + (-x)}\right) dx = \int_{-1}^{1} \log\left(\frac{2 + x}{2 - x}\right) dx \]

### Step 2: Combine the two integrals

Now, we have two expressions for \( I \):

1. \( I = \int_{-1}^{1} \log\left(\frac{2-x}{2+x}\right) dx \)
2. \( I = \int_{-1}^{1} \log\left(\frac{2+x}{2-x}\right) dx \)

Adding these two integrals gives:

\[ 2I = \int_{-1}^{1} \log\left(\frac{2-x}{2+x}\right) + \log\left(\frac{2+x}{2-x}\right) dx \]

### Step 3: Simplify the logarithmic expression

Using the property of logarithms that states \( \log a + \log b = \log(ab) \):

\[ 2I = \int_{-1}^{1} \log\left(\left(\frac{2-x}{2+x}\right) \cdot \left(\frac{2+x}{2-x}\right)\right) dx \]

Now, simplifying the expression inside the logarithm:

\[ \frac{2-x}{2+x} \cdot \frac{2+x}{2-x} = 1 \]

Thus, we have:

\[ 2I = \int_{-1}^{1} \log(1) \, dx \]

### Step 4: Evaluate the integral

Since \( \log(1) = 0 \):

\[ 2I = \int_{-1}^{1} 0 \, dx = 0 \]

### Step 5: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = 0 \]

### Final Answer

The value of the integral is:

\[ \boxed{0} \]