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(A) The integral of an odd function over a symmetric interval $[-a, a]$ is zero.Let $f(x) = \log \left( \frac{2+x}{2-x} \right)$.Then $f(-x) = \log \l

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$\log \left( \frac{2+x}{2-x} \right)$

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(A) The integral of an odd function over a symmetric interval $[-a, a]$ is zero.Let $f(x) = \log \left( \frac{2+x}{2-x} \right)$.Then $f(-x) = \log \left( \frac{2-x}{2-(-x)} \right) = \log \left( \frac{2-x}{2+x} \right) = \log \left( \left( \frac{2+x}{2-x} \right)^{-1} \right) = -\log \left( \frac{2+x}{2-x} \right) = -f(x)$.Since $f(-x) = -f(x)$, the function is odd, and therefore $\int_{-2}^{2} f(x) dx = 0$.

One of the possible functions $f(x)$ which satisfies $\int_{-2}^{2} f(x) dx = 0$ is

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