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Q.

The value of f∫1+xxdx, is

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a

21+x+log⁡1+x−11+x+1+C

b

21+x+C

c

loge⁡1+x−11+x+1+C

d

1+x−11+x+1+C

answer is A.

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Detailed Solution

Let I=∫1+xxdx⇒ I=∫t2t2−1−2tdt, where 1+x=t2⇒ t=2∫t2−1+1t2−1dt=2∫1+1t2−1dt⇒ I=2t+log⁡t−1t+1+C=21+x+log⁡1+x−11+x+1+C
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