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Q.

∫x−1xx+1dx is equal to

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a

ln⁡x−x2−1−tan−1⁡x+C

b

ln⁡x+x2−1−tan−1⁡x+C

c

ln⁡x−x2−1−sec−1⁡x+C

d

ln⁡x+x2−1−sec−1⁡x+C

answer is D.

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Detailed Solution

I=∫x−1xx+1dx=∫x−1xx2−1=∫dxx2−1−∫dxxx2−1=ln⁡x+x2−1−sec−1⁡x+C
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