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I'm learning single variable calculus right now. Right now trying to understand integration with partial fraction. I'm confused in a problem from sometime. I think I'm doing right but answer in my book is something else. Please have a look at the images.Image 1

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The answer given in my book is $\frac{x}{2}+ \log x - \frac{3}{4}\log(1-2x) + C$
Please help. thank you in advance.

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Actually $\ln2x=\ln2+\ln x$

So, $\ln2x+C=\ln x+(\ln2+C)=\ln x+C'$

Using de facto partial fraction decomposition,

$$\dfrac{1-x^2}{x(1-2x)}=A+\dfrac Bx+\dfrac C{1-2x}$$

$$\iff1-x^2=Ax(1-2x)+B(1-2x)+Cx=-2Ax^2+x(A+C-2B)+B$$

Comparing the constants $B=1$

Comparing the coefficients of $x^2,-1=-2A$

Compare the coefficients of $x$

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  • $\begingroup$ Partial Fractions approach is the key concept to solve this problem. $\endgroup$ – NoChance Aug 23 '15 at 6:24
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$\log2x=\log2+\log x$ and $\log2$ is absorbed in $C$. So your answer agrees with what is in the book.

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