Question

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

Answer 1

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

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

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

Substituting $x=0$ and $\frac{1}{2}$ in equation (1), we obtain
$A=2$ and $B=3$

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

Substituting in equation (1), we obtain

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

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

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

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