Question

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

• A. $-\frac{1}{2}\log |1-x|-\frac{1}{6}\log |1+x|+\frac{2}{3}\log |1-2x|+c$
• B. $\frac{1}{2}\log |1-x|-\frac{1}{6}\log |1+x|+\frac{2}{3}\log |1-2x|+c$
• C. $\frac{1}{2}\log |1x|+\frac{1}{6}\log |1+x|+\frac{2}{3}\log |12x|+c$
• D. $\frac{1}{2}\log |1x|+\frac{1}{6}\log |1+x|\frac{2}{3}\log |1-2x|+c$

Answer 1

The correct option is A $-\frac{1}{2}\log |1-x|-\frac{1}{6}\log |1+x|+\frac{2}{3}\log |1-2x|+c$

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

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

Solving by the method of partial fractions,

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

$\Rightarrow 1=A(1-2x)(x+1)+B(x+1)(x-1)+C(x-1)(1-2x)$

Put $x=1\Rightarrow -2A=1\therefore A=-\frac{1}{2}$

Put $x=-1\Rightarrow 1=C(-1-1)(1+2)\Rightarrow -6C=1$

$\therefore C=-\frac{1}{6}$

Put $x=0\Rightarrow A-B-C=1\Rightarrow -\frac{1}{2}-B+\frac{1}{6}=1$

$\Rightarrow -B=1+\frac{1}{2}-\frac{1}{6}=\frac{3}{2}-\frac{1}{6}=\frac{9-1}{6}=\frac{8}{6}$

$\therefore B=-\frac{4}{3}$

$\therefore A=-\frac{1}{2},B=-\frac{4}{3},C=-\frac{1}{6}$

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

$=-\frac{1}{2}\int \frac{dx}{x-1}-\frac{4}{3}\int \frac{dx}{1-2x}-\frac{1}{6}\int \frac{dx}{x+1}$

$=-\frac{1}{2}\log |x-1|-\frac{4}{3}\times \frac{-1}{2}\log |1-2x|-\frac{1}{6}\log |x+1|+c$

$=-\frac{1}{2}\log |x-1|+\frac{2}{3}\log |1-2x|-\frac{1}{6}\log |x+1|+c$