Question

Evaluate the integral: \[ \int_{-1}^{1} \log\left(\frac{2 - x}{2 + x}\right) dx \]

• A. \( 2 \log\left(\frac{1}{2}\right) \)
• B. \( \log\left(\frac{3}{4}\right) \)
• C. \( 0 \)
• D. \( \log 2 \)

Hint:

For definite integrals over symmetric intervals \([-a, a]\), check whether the integrand is even or odd. If it’s odd, the integral is always zero.

Answer 1

The Correct Option is C

Let: \[ I = \int_{-1}^{1} \log\left(\frac{2 - x}{2 + x}\right) dx \] We’ll use the property of definite integrals: \[ \int_{-a}^{a} f(x) dx = \begin{cases} 2\int_{0}^{a} f(x) dx, & \text{if } f(x) \text{ is even}
0, & \text{if } f(x) \text{ is odd} \end{cases} \] Let us test the symmetry of the function: Define \[ f(x) = \log\left(\frac{2 - x}{2 + x}\right) \] Now compute \( f(-x) \): \[ f(-x) = \log\left(\frac{2 + x}{2 - x}\right) = -\log\left(\frac{2 - x}{2 + x}\right) = -f(x) \] So, \( f(x) \) is odd. Therefore: \[ \int_{-1}^{1} f(x) dx = 0 \] Final Answer: \[ \boxed{0} \]