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1x1+xdx is equal to 

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a
−21−x+cos−1⁡x+x+x2+C
b
21−x+cos−1⁡x+x−x2+C
c
−21−x+cos−1⁡x+x−x2+C
d
None of the above

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detailed solution

Correct option is C

Let t=cos−1⁡x Now, ∫1−x1+xdx=∫1−cos⁡t1+cos⁡t⋅[2cos⁡t(−sin⁡t)]dt⇒  I=−2∫2sin2⁡t/22cos2⁡t/2⋅2sin⁡t2⋅cos⁡t2⋅cos⁡tdt∵1−cos⁡x=2sin2⁡x2,1+cos⁡x=2cos2⁡x2and sin⁡2t=2sin⁡tcos⁡t =−4∫sin2⁡t2⋅cos⁡tdt=−4∫1−cos⁡t2⋅cos⁡tdt=−2∫cos⁡t−cos2⁡tdt=−2∫cos⁡t−1+cos⁡2t2dt=−2sin⁡t+t+12sin⁡2t+C=−2sin⁡t+t+12×2sin⁡tcos⁡t+C=−21−cos2⁡t+t+122⋅1−cos2⁡t⋅cos⁡t+C=−21−x+cos−1⁡x+x⋅1−x+C=−21−x+cos−1⁡x+x−x2+C∴∫1−x1+xdx=−21−x+cos−1⁡x+x−x2+C


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