The integral \( \int_{-1 / 2}^{1 / 2}\left([x]+\log \frac{1+x}{1-x}...
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The integral \( \int_{-1 / 2}^{1 / 2}\left([x]+\log \frac{1+x}{1-x}\right) d x \) equals
(a) \( -1 / 2 \)
(b) 0
(c) 1
(d) \( -2 \log 2 \)
\( \mathrm{P} \)
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