Question

If $\int\frac{3x+1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx$ $A \log |x - 1| B \log |x - 2| + C \log |x - 3| + C$ , then the values of $A, B$ and $C$ are respectively

• A. 5, -7, -5
• B. 2, -7, -5
• C. 5, -7, 5
• D. 2, -7, 5

Answer 1

Answer: (D)

Let $\frac{3 x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\dots$(i)
$\Rightarrow \int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x$
$=A \log |x-1|+B \log |x-2|+C \log |x-3|+C$
Now, $3 x+1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ [From eqn. (1)]
Putting $x=1, x=2, x=3$ in the above equation one at a time, we get
$A=2, B=-7, C=5$