$\frac{1-x^2}{x-2 x^2}$ is an improper fraction therefore we convert it into a proper fraction. Divide $1-\mathrm{x}^2$ by $\mathrm{x}-2 \mathrm{x}^2$ by long divison.
$\frac{1-x^2}{x-2 x^2}=\frac{1}{2}\left[\frac{\left(2 x^2-x\right)+(x-2)}{2 x^2-x}\right]=\frac{1}{2}\left[1+\frac{x-2}{2 x^2-x}\right]$
Now, $\frac{x-2}{2 x^2-x}=\frac{x-2}{x(2 x-1)}=\frac{A}{x}+\frac{B}{2 x-1}$
$\Rightarrow \mathrm{x}-2 \equiv \mathrm{A}(2 \mathrm{x}-1)+\mathrm{Bx} \quad \ldots(i)$
Put $x=0, \frac{1}{2}$ in (i), we get : $A=2 B=-3$
$\therefore \int \frac{1-x^2}{x(1-2 x)} d x=\frac{1}{2} x+\int \frac{1}{x} d x+\frac{3}{2} \int \frac{1}{1-2 x} d x$
$=\frac{1}{2} x+\log |x|-\frac{3}{4} \log |1-2 x|+c$