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Q.

∫01 x−1x2−1dx

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a

ℓn⁡(2)

b

ℓn(6) – ℓn(3)

c

ℓn(3) – ℓn(6)

d

– ℓn(2)

answer is A.

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Detailed Solution

∫01 x−1x2−1=∫01 x−1(x+1)(x−1)=∫01 1(x+1)dx=ℓn(x+1)01=ℓn2
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