Question

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

Answer 1

Consider the given integral.

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

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

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

$1-x^2=A-2Ax+Bx$

Here,

$A=1$

$0=-2A+B$

$B=2$

Therefore,

$I=\int \frac{1}{x}dx+\int \frac{2}{1-2x}dx$

$I=\int \frac{1}{x}dx-\int \frac{1}{x-\frac{1}{2}}dx$

$I=\log x-\log (x-\frac{1}{2})+C$

$I=\log \left(\frac{2x}{2x-1}\right)+C$

Hence, this is the answer.