Question

$\underset{0}{\overset{1}{\int }}\frac{x}{x+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{x}{x+1}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_0^{1}1-\frac{1}{x+1}\mathit{d}x \\ \Rightarrow I={\left[x-\log \left(x+1\right)\right]}_0^1 \\ \Rightarrow I=1-\log 2-(0-\log 1) \\ \Rightarrow I=\log e-\log 2 \\ \Rightarrow I=\log \frac{e}{2} \end{gathered}$$