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Q.

∫01tan−12x−11+x−x2dx=

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a

π4

b

ln2

c

π4−ln2

d

0

answer is D.

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Detailed Solution

I=∫01tan−12x−11+x−x2dx,x→1−x I=∫01tan−11−2x1+x−x2dx 2I=0→I=0
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