Question

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

• A. $\log x+\frac{3}{4}\log (1-2x)+c$
  
• B. $\log x-\frac{3}{4}\log (1-2x)+c$
• C. $\log x-\frac{3}{4}\log (1+2x)+c$
  
• D. $\log x+\frac{3}{4}\log (1+2x)+c$
  

Answer 1

The correct option is B $\log x-\frac{3}{4}\log (1-2x)+c$

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

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

Put $x=0\Rightarrow A=1$

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

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

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

$=\log x-\frac{3}{4}\log (1-2x)+c$