If ∫(2x−1)dxx4−2×3+x+1=Atan−1⁡f(x)3+C then 

 If (2x1)dxx42x3+x+1=Atan1f(x)3+C then 

  1. A

    A=13

  2. B

    A=2

  3. C

    f(x)=2x22x1

  4. D

    f(x)=2x2+2x+1

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    Solution:

    I=(2x1)dxx2x2x2x+1 Put x2x=tI=dtt2t+1=dtt122+34I=dtt122+322I=23tan12x22x13+C

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