Question

Evaluate: $\int \frac{1-x^2}{x(1-2x)}dx$.

Answer 1

$\int \frac{1-x^2}{x(1-2x)}dx$

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

By partial fraction $A=1$

$B=2$

$\int \frac{1}{x}dx-\int \frac{2}{1-2x}dx-\int \frac{x}{1-2x}dx$

$=\ln x-\frac{2\ln (1-2x)}{-2}+\frac{1}{2}\int \frac{1-2x+1}{1-2x}dx$

$=\ln x+\ln (1-2x)+\frac{1}{2}\left[\int 1+\frac{1}{1-2x}dx\right]$

$=\ln x+\ln (1-2x)+\frac{x}{2}+\frac{1}{2}.\frac{\ln (1-2x)}{-2}$

$\ln x+\frac{3}{4}\ln (1-2x)+\frac{x}{2}$