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 Evaluate 2xx2+1x2+2dx

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a
log⁡x2+1−log⁡x2+2+C
b
log⁡x2+2−log⁡x2+1+C
c
log⁡x2+1+log⁡x2+2+C
d
None of these

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detailed solution

Correct option is A

Let I=∫2xx2+1x2+2dx  Putting x2=t and 2xdx=dt, we get  I=∫dt(t+1)(t+2)=∫1t+1−1t+2dt=log⁡|t+1|−log⁡|t+2|+C=log⁡x2+1−log⁡x2+2+C


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