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$\int_{-1 / 2}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x=$
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$0$
$\int_{\frac{-1}{2}}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x=0\left[\because \int_{-a}^a f(x) \mathrm{d} x=0\right.$ if $f(x)$ is odd $]$
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