Question

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

Answer 1

$\begin{array}{cc}\int \frac{1-x^2}{x\left(1-2x\right)}dx=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x\left(1-2x\right)}\right)& [\because Dividing\left(1-x^2\right)byx\left(1-2x\right)]\\ \frac{2-x}{x\left(1-2x\right)}=\frac{A}{x}+\frac{B}{\left(1-2x\right)}& \\ \left(2-x\right)=A\left(1-2x\right)+Bx\to \left(i\right)& \\ Putx=0inequation\left(i\right)& \\ \frac{2-x}{x\left(1-2x\right)}=\frac{2}{x}+\frac{3}{1-2x}& \end{array}$

Substituting in equation (i), we get

$\begin{array}{c}\frac{1-x^2}{x\left(1-2x\right)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2}{x}+\frac{3}{\left(1-2x\right)}\right\}\\ =\frac{x}{2}+\int \frac{1}{2}\left(\frac{2}{x}+\frac{3}{1-2x}\right)dx\\ =\frac{x}{2}+\log \left|x\right|+\frac{3}{2\left(-2\right)}\log \left|1-2x\right|+c\\ =\frac{x}{2}+\log \left|x\right|-\frac{3}{4}\log \left|1-2x\right|+c\end{array}$