$\int_{2}^{3} \frac{d x}{x^{2}+x}=$

Question

Aicharya · Accepted Answer

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

$\int_{2}^{3} \frac{d x}{x^{2}+x}=$

Solution