Question
Integrate the rational function $\frac{1-x^{2}}{x(1-2 x)}$

Answer

On dividing $1 - x^2$ by x(1 - 2x), we get,$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)$.......(i)
Now, let $\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}$
(2 - x) = A(1 - 2x) + Bx …...(ii)
Now, substituting x = 0 and $\frac{1}{2}$ in equation (ii), we get,
A = 2 and B = 3
Thus, $\frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{(1-2 x)}$
Now, putting this value in equation (ii), we get,
$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{(1-2 x)}\right)$
$\Rightarrow$$\int \frac{1-x^{2}}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{(1-2 x)}\right)\right\} d x$
= $\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C$
= $\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C$

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