Question

$\int \frac{2x-1}{(x-1)(x+2)(x-3)}dx=A\log |x-1|+B\log |x+2|+C\log |x-3|+K$. Then $A,B,C$ are respectively

• A. $\frac{1}{6},\frac{-1}{3},\frac{1}{3}$
• B. $\frac{-1}{6},\frac{1}{3},\frac{1}{3}$
• C. $\frac{-1}{6},\frac{-1}{3},\frac{1}{2}$
• D. $\frac{1}{6},\frac{1}{3},\frac{1}{5}$

Answer 1

The correct option is C $\frac{-1}{6},\frac{-1}{3},\frac{1}{2}$

We will use partial fractions to solve the integral.
$\text{Let}\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x+2}+\frac{R}{x-3}$
$\text{Then},2x-1=P(x+2)(x-3)+Q(x-1)(x-3)+R(x-1)(x+2)$
$\text{Put}x=1$, we get $P=\frac{-1}{6}$
$\text{Put}x=3$, we get $R=\frac{1}{2}$
$\text{Put}x=-2$, we get $Q=\frac{-1}{3}$
$\therefore \int \frac{2x-1}{(x-1)(x+2)(x-3)}dx$
$=\int (\frac{-1}{6(x-1)}+\frac{-1}{3(x+2)}+\frac{1}{2(x-3)})dx$
$=\frac{-1}{6}\log |x-1|+\left(\frac{-1}{3}\right)\log |x+2|+\frac{1}{2}\log |x-3|+K$
On comparing,
$A=-\frac{1}{6},B=-\frac{1}{3}andC=\frac{1}{2}$