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Q.

The value of ∫1+log⁡xxx2−1dx is

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a

sec−1xx+c

b

tan−1xx+c

c

logxx+x2x−1+c

d

cot−1xx+c

answer is A.

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Detailed Solution

I=∫xx(1+log⁡x)xxxx−1 Let xx=t∴xlog⁡x=log⁡t⇒(log⁡x+1)dx=1tdt⇒xx(1+log⁡x)dx=dt∴I=∫1tt2−1dt=sec−1⁡(t)=sec−1⁡xx+C
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