Question

Evaluate \( \int_{-1}^{1} \log \left( \frac{2 + x}{2 - x} \right) \, dx \)

Hint:

When evaluating integrals of symmetric functions, try to exploit the symmetry by substituting \( x = -t \) to simplify the problem.

Answer 1

Step 1: Symmetry of the integrand.
Let us observe that the integrand is symmetric with respect to the limits \( -1 \) and \( 1 \). Specifically, if we substitute \( x = -t \) into the integral, the integrand remains unchanged. Thus, we can simplify the evaluation by considering the symmetry and the fact that the function is odd. Step 2: Breaking the integral into simpler parts.
Using properties of logarithms, we can rewrite the expression inside the logarithm: \[ \log \left( \frac{2 + x}{2 - x} \right) = \log(2 + x) - \log(2 - x) \] Step 3: Solving the integral.
Now, split the integral into two parts: \[ \int_{-1}^{1} \log \left( \frac{2 + x}{2 - x} \right) \, dx = \int_{-1}^{1} \log(2 + x) \, dx - \int_{-1}^{1} \log(2 - x) \, dx \] Both integrals can be evaluated using standard integration techniques. After solving the integrals, we find that the result is 0. Conclusion:
Thus, the value of the integral is 0.