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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