Evaluate `int_(-1)^(1) log ((2-x)/( 2+x)) dx `

Explanation

To evaluate 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:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(-x) \, dx \]

### Step 1: Define the integral
Let \( I = \int_{-1}^{1} \log \left( \frac{2-x}{2+x} \right) dx \).
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