$\int \frac{1-{\mathrm{x}}^2}{\mathrm{x}(1-2\mathrm{x})}\mathrm{dx}$ is equal to

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

Here, degree of numerator is equal to degree of denominator, 

so divide the numerator by denominator. 

Thus,

$\begin{array}{l}\frac{1-{\mathrm{x}}^2}{\mathrm{x}(1-2\mathrm{x})}=\frac{{\mathrm{x}}^2-1}{2{\mathrm{x}}^2-\mathrm{x}}=\frac{1}{2}+\frac{\frac{1}{2}\mathrm{x}-1}{2{\mathrm{x}}^2-\mathrm{x}}\\ \mathrm{I}=\int \frac{1}{2}\mathrm{dx}+\int \frac{\frac{1}{2}\mathrm{x}-1}{\left(2{\mathrm{x}}^2-\mathrm{x}\right)}\mathrm{dx}\Rightarrow \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2---\left(\mathrm{i}\right)\end{array}$

$\begin{array}{l}\mathrm{where} {\mathrm{l}}_1=\int \frac{1}{2}\mathrm{dx}\text{ and }{\mathrm{I}}_2=\int \frac{\frac{1}{2}\mathrm{x}-1}{\left(2{\mathrm{x}}^2-\mathrm{x}\right)}\mathrm{dx}\\ \mathrm{now}, {\mathrm{l}}_1=\int \frac{1}{2}\mathrm{dx}=\frac{1}{2}\mathrm{x}+{\mathrm{C}}_1\text{ and }{\mathrm{I}}_2=\int \frac{\frac{1}{2}\mathrm{x}-1}{2{\mathrm{x}}^2-\mathrm{x}}\mathrm{dx}\\ \mathrm{Let} \frac{\frac{1}{2}\mathrm{x}-1}{\left(2{\mathrm{x}}^2-\mathrm{x}\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{(2\mathrm{x}-1)}\Rightarrow \frac{\frac{1}{2}\mathrm{x}-1}{\left(2{\mathrm{x}}^2-\mathrm{x}\right)}=\frac{\mathrm{A}(2\mathrm{x}-1)+\mathrm{Bx}}{\mathrm{x}(2\mathrm{x}-1)}\\ \Rightarrow \frac{1}{2}\mathrm{x}-1=2\mathrm{Ax}-\mathrm{A}+\mathrm{Bx}\\ \Rightarrow \frac{1}{2}\mathrm{x}-1=\mathrm{x}(2\mathrm{A}+\mathrm{B})-\mathrm{A}\end{array}$

On comparing the coefficient of x and constant term on both sides, we get

$\begin{array}{r}2\mathrm{A}+\mathrm{B}=\frac{1}{2}-----\left(\mathrm{ii}\right)\\ -\mathrm{A}=-1\Rightarrow \mathrm{A}=1----\left(\mathrm{iii}\right)\end{array}$

From Eq. (iii), $2\times 1+\mathrm{B}=\frac{1}{2}$

$\begin{array}{l}\Rightarrow \mathrm{B}=\frac{1}{2}-2=\frac{-3}{2}\\ \therefore {\mathrm{I}}_2=\int \left[\frac{1}{\mathrm{x}}-\frac{3}{2(2\mathrm{x}-1)}\right]\mathrm{dx}\\ =\int \frac{1}{\mathrm{x}}\mathrm{dx}-\frac{3}{2}\int \frac{1}{2\mathrm{x}-1}\mathrm{dx}\\ \Rightarrow {\mathrm{I}}_2=\log \mathrm{x}-\frac{3}{2}\frac{\log |2\mathrm{x}-1|}{2}+{\mathrm{C}}_2\end{array}$

On putting the values of l₁ and I₂ in Eq. (i), we get

$\text{I=}\frac{1}{2}\mathrm{x}+\log \mathrm{x}-\frac{3}{4}\log |2\mathrm{x}-1|+\mathrm{C}\left({\mathrm{C}}_1+{\mathrm{C}}_2=\mathrm{C}\right)$