Answer
$$-\dfrac{5}{6}\ln|3x+1|+\dfrac{1}{2}\ln |x-1|+C $$
Work Step by Step
Step #1: Complete Partial Fraction Decomposition of the integrand.
$$\dfrac{3-x}{3x^2-2x-1}=\dfrac{3-x}{(3x+1)(x-1)}=\dfrac{A}{3x+1}+\dfrac{B}{x-1}$$
Step #2: Multiply by the LCD of $\color{blue}{(3x+1)(x-1)}$ and simply by collecting like terms.
\begin{align*}
3-x&=A(x-1)+B(3x+1)\\
3-x&=Ax-A+3Bx+B\\
3-x&=(A+3B)x+(-A+B)
\end{align*}
Step #3: Match the coefficients on both sides of the equation to get a system of equations. Solve for $A$ and $B$.
\begin{align*}
&\Rightarrow \left\{\begin{array}{l}{\text{ }A+3 B=-1} \\ {-A+B=3}\end{array}\right.\\
\\
&\Rightarrow\left\{\begin{array}{l}{\text{ }A+3B=-1} \\ {B=3+A}\end{array}\right.\\
\\
&\Rightarrow\left\{\begin{array}{l}{\text{ }A+3(3+A)=-1} \\ {B=3+A}\end{array}\right.\\
\\
&\Rightarrow\left\{\begin{array}{l}{\text{ }4A+9=-1} \\ {B=3+A}\end{array}\right.\\
\\
&\Rightarrow\left\{\begin{array}{l}{\text{ }A=-\dfrac{5}{2}} \\ {B=3+A}\end{array}\right.\\
\\
&\Rightarrow\left\{\begin{array}{l}{\text{ }A=-\dfrac{5}{2}} \\ {B=\dfrac{1}{2}}\end{array}\right.
\end{align*}
Step #4: Use the values for $A$ and $B$ to rewrite the integral
$$\int \dfrac{3-x}{3x^2-2x-1}dx=\int\left( \dfrac{-{}^{5}{\mskip -5mu/\mskip -3mu}_{2}}{3x+1}+\dfrac{{}^{1}{\mskip -5mu/\mskip -3mu}_{2}}{x-1}\right)dx$$
Step #5: Integrate
\begin{align*}
\int\left( \dfrac{-{}^{5}{\mskip -5mu/\mskip -3mu}_{2}}{3x+1}+\dfrac{{}^{1}{\mskip -5mu/\mskip -3mu}_{2}}{x-1}\right)&=\int \dfrac{-{}^{5}{\mskip -5mu/\mskip -3mu}_{2}}{3x+1}dx+\int \dfrac{{}^{1}{\mskip -5mu/\mskip -3mu}_{2}}{x-1}dx\\
&=-\dfrac{5}{2}\int\dfrac{1}{3x+1}dx+\dfrac{1}{2}\int\dfrac{1}{x-1}dx\\
&=-\dfrac{5}{6}\int\dfrac{3}{3x+1}dx+\dfrac{1}{2}\int\dfrac{1}{x-1}dx\\
\\
&=\boxed{-\dfrac{5}{6}\ln|3x+1|+\dfrac{1}{2}\ln|x-1|+C}\\
\end{align*}
