Answer
$-\frac{1}{2}-\ln2-\frac{1}{3}\ln5$
Work Step by Step
$\int_2^3\frac{x(3-5x)}{(3x-1)(x-1)^2}dx$ = $\int_2^3[\frac{3x-5x^2}{(3x-1)(x-1)^2}]dx$ = $\int_2^3\left[\frac{A}{3x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\right]dx$
$3x-5x^2$ = $A(x-1)^2+B(3x-1)(x-1)+C(3x-1)$
$3x-5x^2$ = $A(x^2-2x+1)+B(3x^2-4x+1)+C(3x-1)$
$-5$ = $A+3B$
$3$ = $-2A-4B+3C$
$0$ = $A+B-C$
$A$ = $1$, $B$ = $-2$, $C$ = $-1$
so
$\int_2^3\frac{x(3-5x)}{(3x-1)(x-1)^2}dx$ = $\int_2^3\left[\frac{1}{3x-1}-\frac{2}{x-1}-\frac{1}{(x-1)^2}\right]dx$ = $\left[\frac{1}{3}\ln|3x-1|-2\ln|x-1|+\frac{1}{x-1}\right]_2^3$ = $\left(\frac{1}{3}\ln8-2\ln2+\frac{1}{2}\right)-\left(\frac{1}{3}\ln5-2\ln1+1\right)$ = $-\frac{1}{2}-\ln2-\frac{1}{3}\ln5$
