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

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

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(C) Let $f(x) = \log \left( \frac{1+x}{1-x} \right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left( \frac{1+(-x)}{1-(-x)} \right) = \log \left( \frac{1-x}{1+x} \right) = \log \left( \left( \frac{1+x}{1-x} \right)^{-1} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.Therefore,$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = 0$.

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

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