Question

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

Answer 1

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

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

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

Put $x=0\Rightarrow 1=A$ or $A=1$

Put $x=\frac{1}{2}\Rightarrow 1-\frac{1}{4}=\frac{1}{2}B$

$\Rightarrow \frac{1}{2}B=\frac{3}{4}$

$\Rightarrow B=\frac{3}{2}$

$\therefore A=1,B=\frac{3}{2}$

Now,$\int \frac{1-x^2}{x(1-2x)}dx=\int \frac{dx}{x}+\frac{3}{2}\int \frac{dx}{1-2x}$

$=\log \left|x\right|-\frac{3}{2}\times \frac{1}{2}\log |1-2x|+c$

$=\log \left|x\right|-\frac{3}{4}\log |1-2x|+c$