The integral is equal to (where

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Problem

The integral ∫−1/21/2{[x]+log⁡e(1+x1−x)}dx\int_{-1/2}^{1/2} \left\{ [x] + \log_e \left( \frac{1+x}{1-x} \right) \right\} dx is equal to (where, [.][. ] denotes greatest integer function) (1) −12-\frac{1}{2} (2) 00 (3) 11 (4) 2log⁡(12)2 \log \left( \frac{1}{2} \right)

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