If ∫(2x−1)dxx4−2x3+x+1=Atan−1f(x)3+C then
A=13
A=2
f(x)=2x2−2x−1
f(x)=2x2+2x+1
I=∫(2x−1)dxx2−x2−x2−x+1 Put x2−x=tI=∫dtt2−t+1=∫dtt−122+34I=∫dtt−122+322⇒I=23tan−12x2−2x−13+C