∫1−x2x(1−2x)dx is equal to

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x+log⁡x−34log⁡|x−1|+C

x2+log⁡x+14log⁡|x−1|+C

x2+log⁡x−34log⁡|2x−1|+C

None of the above

answer is C.

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Detailed Solution

Let ∫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−xI=∫12dx+∫12x−12x2−xdx⇒I=I1+I2---iwhere l1=∫12dx and I2=∫12x−12x2−xdxnow, l1=∫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 get2A+B=12-----ii−A=−1⇒A=1----iiiFrom 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

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