int_{ - 1}^1 {log left( frac{2 - x}{2 + x} ri | vedclass

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(D) Let $f(x) = \log \left( \frac{2 - x}{2 + x} \right)$.Now,calculate $f(-x)$:$f(-x) = \log \left( \frac{2 - (-x)}{2 + (-x)} \right) = \log \left( \f

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(D) Let $f(x) = \log \left( \frac{2 - x}{2 + x} \right)$.Now,calculate $f(-x)$:$f(-x) = \log \left( \frac{2 - (-x)}{2 + (-x)} \right) = \log \left( \frac{2 + x}{2 - x} \right)$.We know that $\log \left( \frac{1}{a} \right) = -\log(a)$,so:$f(-x) = \log \left( \left( \frac{2 - x}{2 + x} \right)^{-1} \right) = -\log \left( \frac{2 - x}{2 + x} \right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.Therefore,$\int_{-1}^{1} \log \left( \frac{2 - x}{2 + x} \right) \, dx = 0$.

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

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