Question

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

Answer 1

It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain

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

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

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

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

A = 2 and B = 3

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

Substituting in equation (1), we obtain

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

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

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

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