Question

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

Answer 1

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

$x-2x^2)\overline{1-x^2}x/2-x^2-+\_\_\_\_\_\_\_\_\_\_\_\_\_1-x/2$

$\therefore 1-x^2=(x-2x^2)(1-x/2)+1/2$

so, equation $\left(1\right)$ becomes

$\Rightarrow \int \frac{1-x/2}{x(1-2x)}dx---\left(2\right)$

Now, $\int \frac{1-x/2}{x(1-2x)}dx=\frac{A}{x}+\frac{B}{1-2x}$

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

comparing powers and coefficients

$A=1$

$B=3/2$

$\Rightarrow \frac{x}{2}+\ln x+\int \frac{3/2}{1-2x}dx$

multiplying and dividing by $(-2)$

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

$\Rightarrow \frac{x}{2}+\ln |x|-\frac{3}{4}\ln |1-2x|+C$