int_{-1}^1 log left(frac{2-x}{2+x}right) d x= | vedclass

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Question

(C) Let $f(x) = \log \left(\frac{2-x}{2+x}\right)$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \log \left(\frac{2-(-x)}{2+(-x)

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(C) Let $f(x) = \log \left(\frac{2-x}{2+x}\right)$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \log \left(\frac{2-(-x)}{2+(-x)}\right) = \log \left(\frac{2+x}{2-x}\right)$.Using the property $\log(a/b) = -\log(b/a)$,we get:$f(-x) = -\log \left(\frac{2-x}{2+x}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function is an odd function.For an odd function,the property of definite integration states that $\int_{-a}^a f(x) dx = 0$.Therefore,$\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x = 0$.

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

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