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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x$.Consider the function $f(x) = \log \left(\frac{2019-x}

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x$.Consider the function $f(x) = \log \left(\frac{2019-x}{2019+x}\right)$.Check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{2019-(-x)}{2019+(-x)}\right) = \log \left(\frac{2019+x}{2019-x}\right)$.Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:$f(-x) = -\log \left(\frac{2019-x}{2019+x}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.For an odd function,the definite integral over a symmetric interval $[-a, a]$ is always $0$,i.e.,$\int_{-a}^{a} f(x) dx = 0$.Therefore,$I = 0$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x=$ . . . . . . .

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