Question

$\underset{1}{\overset{2}{\int }}\frac{x+3}{x\left(x+2\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_1^2\frac{x+3}{x\left(x+2\right)}\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_1^2\left(\frac{x}{x\left(x+2\right)}+\frac{3}{x\left(x+2\right)}\right)\mathit{d}x \\ \mathit{\Rightarrow }I\mathit{=}{\int }_{\mathit{1}}^{\mathit{2}}\frac{\mathit{d}\mathit{x}}{\left(\mathit{x}\mathit{+}\mathit{2}\right)}\mathit{+}{\int }_{\mathit{1}}^{\mathit{2}}\frac{\mathit{3}}{\mathit{x}\left(\mathit{x}\mathit{+}\mathit{2}\right)}\mathit{}\mathit{d}x \\ \Rightarrow I={\left[\log \left(x+2\right)\right]}_1^2+\frac{3}{2}{\int }_1^2\left(\frac{1}{x}-\frac{1}{x+2}\right)dx \\ \Rightarrow I={\left[\log \left(x+2\right)\right]}_1^2+\frac{3}{2}{\left[\log x-\log \left(x+2\right)\right]}_1^2 \\ \Rightarrow I=\log 4-\log 3+\frac{3}{2}\left[\log 2-\log 4-0+\log 3\right] \\ \Rightarrow I=\log 4-\log 3+\frac{3}{2}\left[-\log 2+\log 3\right] \\ \Rightarrow I=2\log 2-\log 3+\frac{3}{2}\log 3-\frac{3}{2}\log 2 \\ \Rightarrow I=\frac{1}{2}\log 2+\frac{1}{2}\log 3 \\ \Rightarrow I=\frac{1}{2}\left(\log 2+\log 3\right) \\ \Rightarrow I=\frac{1}{2}\log 6 \end{gathered}$$