$\begin{aligned} \mathrm{I} &=\int_{2}^{3} \frac{\mathrm{dx}}{\mathrm{x}^{2}+\mathrm{x}}=\int_{2}^{3} \frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)} \\ \therefore \mathrm{I} &=\int_{2}^{3}\left[\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}\right] \mathrm{dx} \\ &=[\log \mathrm{x}-\log (\mathrm{x}+1)]_{2}^{3}=\left[\log \left(\frac{\mathrm{x}}{\mathrm{x}+1}\right)\right]_{2}^{3} \\ &=\log \left(\frac{3}{4}\right)-\log \left(\frac{2}{3}\right)=\log \left(\frac{3}{4} \times \frac{3}{2}\right)=\log \left(\frac{9}{8}\right) \end{aligned}$