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

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a
A=13
b
A=2
c
f(x)=2x2−2x−1
d
f(x)=2x2+2x+1

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detailed solution

Correct option is C

I=∫(2x−1)dxx2−x2−x2−x+1 Put x2−x=tI=∫dtt2−t+1=∫dtt−122+34I=∫dtt−122+322⇒I=23tan−1⁡2x2−2x−13+C


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