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

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Question

(A) Let $f(x) = \log \left(\frac{1+x}{1-x}\right)$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \log \left(\frac{1+(-x)}{1-(-x)

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

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

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