Q.
The definite integral ∫01 1−x1+xdx is equal to
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a
1
b
π
c
π2−1
d
π2+12
answer is C.
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Detailed Solution
I=∫01 1−x1+xdx=∫01 1−x1−x2dx=sin−1x+1−x201=π2−1
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