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

The given integrand is $\frac{1-x^{2}}{x-2x^{2}}$. Since the degree of the numerator is equal to the degree of the denominator,it is an improper fraction.
First,we perform long division:
$\frac{1-x^{2}}{-2x^{2}+x} = \frac{1}{2} + \frac{2-x}{2x(1-2x)}$
Now,we decompose $\frac{2-x}{x(1-2x)}$ into partial fractions:
$\frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}$
$2-x = A(1-2x) + Bx$
Setting $x=0$,we get $A=2$.
Setting $x=\frac{1}{2}$,we get $2-\frac{1}{2} = B(\frac{1}{2}) \Rightarrow \frac{3}{2} = \frac{B}{2} \Rightarrow B=3$.
Thus,$\frac{1-x^{2}}{x(1-2x)} = \frac{1}{2} + \frac{1}{2} \left( \frac{2}{x} + \frac{3}{1-2x} \right) = \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)}$.
Integrating with respect to $x$:
$\int \left( \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)} \right) dx = \frac{x}{2} + \log|x| + \frac{3}{2} \cdot \frac{\log|1-2x|}{-2} + C$
$= \frac{x}{2} + \log|x| - \frac{3}{4} \log|1-2x| + C$.