Question

Integrate the rational functions.
$\int \frac{1-x^2}{x(1-2x)}dx.$

Answer 1

Let $I=\int \frac{1-x^2}{x(1-2x)}dx$. Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.
Thus, $\frac{1-x^2}{x(1-2x)}=\frac{x^2-1}{2x^2-x}=\frac{1}{2}+\frac{\frac{1}{2}x-1}{2x^2-x}$
$\therefore I=\int \frac{1}{2}dx+\int \frac{\frac{1}{2}x-1}{(2x^2-x)}dx\Rightarrow I=I_1+I_2.......\left(i\right)$
where, $I_1=\int \frac{1}{2}dxand{I}_2=\int \frac{\frac{1}{2}x-1}{(2x^2-x)}dx$
Now, $I_1=\int \frac{1}{2}dx=\frac{1}{2}x+C_1$
and $I_2=\int \frac{\frac{1}{2}x-1}{2x^2-x}dx$
Let $\frac{\frac{1}{2}x-1}{(2x^2-x)}=\frac{A}{x}+\frac{B}{(2x-1)}\Rightarrow \frac{\frac{1}{2}x-1}{(2x^2-x)}=\frac{A(2x-1)+Bx}{x(2x-1)}$
$\Rightarrow \frac{1}{2}x-1=2Ax-A+Bx\Rightarrow \frac{1}{2}x-1=x(2A+B)-A$

$2A+B=\frac{1}{2}.....\left(i\right)$
and $-A=-1\Rightarrow A=1....\left(ii\right)$
From Eq. (i), $2\times 1+B=\frac{1}{2}$
$\Rightarrow B=\frac{1}{2}-2=\frac{-3}{2}\therefore I_2=\int [\frac{1}{x}-\frac{3}{2(2x-1)}]dx=\int \frac{1}{x}dx-\frac{3}{2}\int \frac{1}{2x-1}dx\Rightarrow I_2=logx-\frac{3}{2}\frac{log|2x-1|}{2}+C_2$
On putting the values of $I_1$ and $I_2$ in Eq (i), we get
$I=\frac{1}{2}x+logx-\frac{3}{4}log|2x-1|+C(C_1+C_2=C$ )