∫1−x2x(1−2x)dx is equal to

Let I=∫1−x2x(1−2x)dxHere, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.Thus,1−x2x(1−2x)=x2−12x2−x=12+12x−12x2−x∴      I=∫12dx+∫12x−12x2−xdx⇒I=I1+I2     .....(i)Where,    I1=∫12dx and I2=∫12x−12x2−xdxNow, I1=∫12dx=12x+C1 and I2=∫12x−12x2−xdxLet   12x−12x2−x=Ax+B(2x−1)⇒12x−12x2−x=A(2x−1)+Bxx(2x−1)⇒ 12x−1=2Ax−A+Bx⇒ 12x−1=x(2A+B)−AOn comparing the coefficient of x and constant term on both sides, we get                            2A+B=12           .....(ii) and    −A=−1⇒A=1           .....(iii) From Eq. (iii), 2×1+B=12⇒  B=12−2=−32∴ I2=∫1x−32(2x−1)dx =∫1xdx−32∫12x−1dx⇒ I2=log⁡x−32log⁡|2x−1|2+C2On putting the values of l1 and I2 in Eq. (i), we getI=12x+log⁡x−34log⁡|2x−1|+C C1+C2=C