Question

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

Answer 1

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