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The integral $\int_{- 1 / 2}^{1 / 2} ([x] + 2 \log \frac{1 + x}{1 - x}) d x$ equal ((where [x] is greatest integer function)

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2 log (1/2)

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Correct option is A

∫−1/21/2 [x]+2log⁡ 1+x1−xdx=∫−1/21/2 [x]dx(since log 1+x1−x is an odd function)=∫−1/20 [x]dx+∫01/2 [x]dx=∫−1/20 (−1)dx=−12.

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The integral ∫−1/21/2 [x]+2log⁡ 1+x1−xdx equal ((where [x] is greatest integer function)

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The integral $\int_{- 1 / 2}^{1 / 2} ([x] + 2 \log \frac{1 + x}{1 - x}) d x$ equal ((where [x] is greatest integer function)