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$\frac{1-x^{2}}{x(1-2 x)}$


It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $\left(1-x^{2}\right)$ by $x(1-2 x)$, we obtain

$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)$

Let $\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}$

$\Rightarrow(2-x)=A(1-2 x)+B x$    ...(1)

Substituting $x=0$ and $\frac{1}{2}$ in equation $(1)$, we obtain

$A=2$ and $B=3$

$\therefore \frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{1-2 x}$

Substituting in equation (1), we obtain

$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2}{x}+\frac{3}{(1-2 x)}\right\}$

$\Rightarrow \int \frac{1-x^{2}}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{1-2 x}\right)\right\} d x$

$=\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+\mathrm{C}$

$=\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+\mathrm{C}$


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