$\begin{aligned} & \text {} \int \frac{2 x+3}{\sqrt{3 x^2-2 x+1}} d x=\frac{1}{3} \int \frac{6 x-2+\frac{11}{3}}{\sqrt{3 x^2-2 x+1}} d x \\ & =\frac{1}{3}\left[\int \frac{(6 x-2) d x}{\sqrt{3 x^2-2 x+1}}+\frac{11}{3} \int \frac{d x}{\sqrt{3 x^2-2 x+1}}\right] \\ & I_1=\int \frac{6 x-2}{\sqrt{3 x^2-2 x+1}} d x \\ & \Rightarrow \sqrt{3 x^2-2 x+1}=t\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{(6 x-2) d x}{2 \sqrt{3 x^2-2 x+1}}=d t \\ & \therefore \quad \int 2 d t=2 t+C=2 \sqrt{3 x^2-2 x+1}+C \\ & I_2=\int \frac{d x}{\sqrt{3 x^2-2 x+1}}=\int \frac{d x}{\sqrt{3\left(x^2-\frac{2}{3} x+\frac{1}{3}\right)}} \\ & =\int \frac{d x}{\sqrt{3}\left[\left(x-\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2\right]^{\frac{1}{2}}} \\ & =\frac{1}{\sqrt{3}} \times \frac{3}{\sqrt{2}} \sin h^{-1}\left(\frac{x-\frac{1}{3}}{\sqrt{2}}\right)^2+C \\ & I_2=\frac{\sqrt{3}}{\sqrt{2}} \sin h^{-1}\left(\frac{3 x-1}{\sqrt{2}}\right)+C \\ & \therefore \quad I=\frac{2}{3} \sqrt{3 x^2-2 x+1}+\frac{11}{9} \times \frac{\sqrt{3}}{\sqrt{2}} \sin h^{-1}\left(\frac{3 x-1}{\sqrt{2}}\right)+C . \\ & =\frac{2}{3} \sqrt{3 x^2-2 x+1}+\frac{11}{3 \sqrt{6}} \sin h^{-1}\left(\frac{3 x-1}{\sqrt{2}}\right)+C\end{aligned}$