Q.

∫−1212([x]+ln(1+x1−x))dx=

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a

−12

b

0

c

1

d

2ln12

answer is A.

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Detailed Solution

Since ln(1+x1−x)  is odd, I=∫−1212[x]dx=∫−120(−1)dx+∫0120.dx =−12
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∫−1212([x]+ln(1+x1−x))dx=