$\frac{1-x^{2}}{x(1-2 x)}$
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
$\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2}{x}+\frac{3}{(1-2 x)}\right\}$
$\Rightarrow \int \frac{1-x^{2}}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{1-2 x}\right)\right\} d x$
$=\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+\mathrm{C}$
$=\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+\mathrm{C}$