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−1⁡x+1−x201=π2−1
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