int_{-4}^{4} log left(frac{8-x}{8+x}right) d | vedclass

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(D) Let $f(x) = \log \left(\frac{8-x}{8+x}\right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{8-(-x)}{8+(

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(D) Let $f(x) = \log \left(\frac{8-x}{8+x}\right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{8-(-x)}{8+(-x)}\right) = \log \left(\frac{8+x}{8-x}\right)$.Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:$f(-x) = -\log \left(\frac{8-x}{8+x}\right) = -f(x)$.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 $\int_{-a}^{a} f(x) d x = 0$.Therefore,$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x = 0$.

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

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