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 Evaluate 1+xxdx

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a
x2+x+log⁡x+12+x2+x+C
b
x2+x+12log⁡x+12-x2+x+C
c
x2+x+12log⁡x+12+x2+x+C
d
None of these

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detailed solution

Correct option is C

∫1+xxdx=∫1+xx1+x1+xdx=∫1+xx2+xdx=12∫2x+1x2+xdx+12∫1x2+xdx=12∫1tdt+12∫1x+122−122dx, where t=x2+x=t+12log⁡x+12+x2+x+C=x2+x+12log⁡x+12+x2+x+C


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