Question

# The value of the integral ∫1−x2x(1−2x)dx is (where C is an arbitrary constant)

A
x2+ln|x|34ln|12x|+C
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B
x4+ln|x|43ln|12x|+C
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C
x2+ln|x|+43ln|12x|+C
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D
x4ln|x|+34ln|12x|+C
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Solution

## The correct option is A x2+ln|x|−34ln|1−2x|+C It can be seen that the given integrand is not a proper rational expression, therefore 1−x2x(1−2x)=1−x2x−2x2=12[(x−2x2)+(2−x)x−2x2]=12+12[2−xx(1−2x)] Let 2−xx(1−2x)=Ax+B(1−2x)⇒(2−x)=A(1−2x)+Bx⋯(1) Equating the coefficients of x and constant term, we obtain −2A+B=−1A=2⇒B=3 Therefore, 2−xx(1−2x)=2x+31−2x Substituting in equation (1), we obtain 1−x2x(1−2x)=12+12[2x+3(1−2x)]⇒∫1−x2x(1−2x)dx =∫[12+12(2x+3(1−2x))]dx =x2+ln|x|+32(−2)ln|1−2x|+C =x2+ln|x|−34ln|1−2x|+C Where C is an arbitary constant.

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