Question

The value of $\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx$ is
(where $m$ is an arbitrary constant)

• A. $\ln |x-1|-2\ln |x-2|+3\ln |x-3|+m$
• B. $\ln |x-1|-3\ln |x-2|+4\ln |x-3|+m$
• C. $\ln |x-1|-5\ln |x-2|+4\ln |x-3|+m$
• D. $\ln |x-1|-\ln |x-2|+4\ln |x-3|+m$

Answer 1

The correct option is C $\ln |x-1|-5\ln |x-2|+4\ln |x-3|+m$

$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\Rightarrow 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\cdots \left(1\right)$
By putting $x=1;x=2$ and $x=3$, we obtain
$A=1,B=-5,C=4\therefore \frac{3x-1}{(x-1)(x+2)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\Rightarrow \int \frac{3x-1}{(x-1)(x-2)(x-3)}dx=\int \{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\}dx=\ln |x-1|-5\ln |x-2|+4\ln |x-3|+m$
Where $m$ is an arbitrary constant.