Question

The value of the integral $\int \frac{1-x^2}{x(1-2x)}dx$ is
(where $C$ is an arbitrary constant)

• A. $\frac{x}{4}+\ln |x|-\frac{4}{3}\ln |1-2x|+C$
• B. $\frac{x}{4}-\ln |x|+\frac{3}{4}\ln |1-2x|+C$
• C. $\frac{x}{2}+\ln |x|+\frac{4}{3}\ln |1-2x|+C$
• D. $\frac{x}{2}+\ln |x|-\frac{3}{4}\ln |1-2x|+C$

Answer 1

The correct option is D $\frac{x}{2}+\ln |x|-\frac{3}{4}\ln |1-2x|+C$

It can be seen that the given integrand is not a proper rational expression, therefore
$\frac{1-x^2}{x(1-2x)}=\frac{1-x^2}{x-2x^2}=\frac{1}{2}\left[\frac{(x-2x^2)+(2-x)}{x-2x^2}\right]=\frac{1}{2}+\frac{1}{2}\left[\frac{2-x}{x(1-2x)}\right]$

Let
$\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\Rightarrow (2-x)=A(1-2x)+Bx\cdots \left(1\right)$
Equating the coefficients of $x$ and constant term, we obtain
$-2A+B=-1A=2\Rightarrow B=3$

Therefore,
$\frac{2-x}{x(1-2x)}=\frac{2}{x}+\frac{3}{1-2x}$
Substituting in equation $\left(1\right)$, we obtain
$\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{1}{2}[\frac{2}{x}+\frac{3}{(1-2x)}]\Rightarrow \int \frac{1-x^2}{x(1-2x)}dx=\int [\frac{1}{2}+\frac{1}{2}(\frac{2}{x}+\frac{3}{(1-2x)})]dx=\frac{x}{2}+\ln |x|+\frac{3}{2(-2)}\ln |1-2x|+C=\frac{x}{2}+\ln |x|-\frac{3}{4}\ln |1-2x|+C$
Where $C$ is an arbitary constant.