Answer
$-\displaystyle \ln 2-\frac{1}{2}-\frac{\ln 5}{3}$
Work Step by Step
$\displaystyle \frac{3x-5x^{2}}{(x-1)^{2}(3x-1)}=\frac{A}{3x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}$
$\displaystyle \frac{(-5x^{2}+3x)(x-1)^{2}(3x-1)}{(x-1)^{2}(3x-1)}=\frac{A(x-1)^{2}(3x-1)}{3x-1}+\frac{B(x-1)^{2}(3x-1)}{x-1}+\frac{C(x-1)^{2}(3x-1)}{(x-1)^{2}}$
$3x-5x^{2}=A(x-1)^{2}+B(3x-1)(x-1)+C(3x-1)$
$3x-5x^{2}=A(x^{2}-2x+1)+B(3x^{2}-4x+1)+3Cx-C$
$-5x^{2}+3x+0=x^{2}(A+3B)+x(-2A-4B+3C)+(A+B-C)$
$\left\{\begin{array}{lllll}
A & +3B & & =-5 & \\
-2A & -4B & +3C & =3 & /+2R_{1}\\
A & +B & -C & =0 & /-R_{1}
\end{array}\right.$
$\left\{\begin{array}{lllll}
A & +3B & & =-5 & \\
& 2B & +3C & =-7 & /\\
& -2B & -C & =5 & /+R_{2}
\end{array}\right.$
$\left\{\begin{array}{llllll}
A & +3B & & =-5 & & \\
& 2B & +3C & =-7 & / & \\
& & 2C & =-2 & \Rightarrow & C=-1
\end{array}\right.$
Eq 2 $\Rightarrow B=-2$
Eq 1 $\Rightarrow A=1$
$\displaystyle \frac{x(3-5x)}{(3x-1)(x-1)^{2}}=\frac{1}{3x-1}-\frac{2}{x-1}-\frac{1}{(x-1)^{2}}$
$\displaystyle \int_{2}^{3}\frac{x(3-5x)}{(3x-1)(x-1)^{2}}dx= \left|\frac{1}{3}\ln|3x-1|-2\ln|x-1|+\frac{1}{x-1}\right|_{2}^{3}$
$=\displaystyle \frac{1}{3}\ln 8-2\ln 2+\frac{1}{2}-(\frac{1}{3}\ln 5-0+1)$
$=\displaystyle \frac{1}{3}\cdot 3\ln 2-2\ln 2+\frac{1}{2}-\frac{1}{3}\ln 5-1$
$=-\displaystyle \ln 2-\frac{1}{2}-\frac{\ln 5}{3}$
