Since $\frac{1-x^2}{x(1-2x)}=\frac{1-x^2}{x-2x^2}$ is an improper fraction, therefore we convert it into a proper fraction by long division method, we get $\frac{x^2-1}{2x^2-x}=\frac{1}{2}+\frac{\frac{x}{2}-1}{2x^2-x}$ $\therefore \int \frac{-1-x^2}{-x+2x^2} \, dx$ $=\frac{1}{2} \int dx + \frac{1}{2} \int \frac{x-2}{2x^2-x} \, dx$ Now, $\frac{x-2}{2x^2-x} = \frac{x-2}{x(2x-1)}$ $= \frac{A}{x} + \frac{B}{2x-1}$ ⇒ x – 2 = A(2x – 1) + Bx ...(i) Putting x = 0 in (i), we get –2 = A(–1) ⇒ A = 2 Putting $x = \frac{1}{2}$ in (i), we get $\frac{1}{2} - 2 = B\left(\frac{1}{2}\right)$ $\Rightarrow 1-4 = B$ $\Rightarrow B = -3$$\therefore \frac{x-2}{2x^2-x}$ $= \frac{2}{x} - \frac{3}{2x-1}$ $= \frac{2}{x} + \frac{3}{1-2x}$$\therefore \int \frac{1-x^2}{x(1-2x)} \, dx$ $= \frac{1}{2} \int dx + \frac{1}{2} \int \left( \frac{2}{x} + \frac{3}{1-2x} \right) dx$$= \frac{1}{2} x + \log \lvert x \rvert - \frac{3}{4} \log \lvert 1-2x \rvert + C$