Put $5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^2\right)+B$
$\Rightarrow 6 \mathrm{~A}=5, \mathrm{~A}=\frac{5}{6}-2=2 \mathrm{~A}+\mathrm{B} \therefore \mathrm{B}=-\frac{11}{3}$
$I=\int \frac{\frac{5}{6}(6 x+2)}{3 x^2+2 x+1} d x-\frac{11}{3} \int \frac{d x}{3 x^2+2 x+1}$
$=\mathrm{I}_1-\frac{11}{3} \mathrm{I}_2 ;$ put $3 \mathrm{x}^2+2 \mathrm{x}+1=\mathrm{t} \therefore(6 \mathrm{x}+2) \mathrm{dx}=\mathrm{dt}$
$\mathrm{I}_1=\frac{5}{6} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{5}{6} \log \mathrm{t}=\frac{5}{6} \log \left(3 \mathrm{x}^2+2 \mathrm{x}+1\right)+\mathrm{c}_1$
and $\mathrm{I}_2=\int \frac{\mathrm{dx}}{3 \mathrm{x}^2+2 \mathrm{x}+1}=\frac{1}{3} \int \frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2}$
$\Rightarrow \mathrm{I}_2=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{3 \mathrm{x}+1}{\sqrt{2}}+\mathrm{c}$
$\therefore \mathrm{I}=\frac{5}{6} \log \left(3 \mathrm{x}^2+2 \mathrm{x}+1\right)-\frac{11}{3} \cdot \frac{1}{\sqrt{2}} \tan ^{-1} \frac{3 x+1}{\sqrt{2}}+\mathrm{c}$