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

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

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

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

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