Question

The value of the integral $\underset{-1}{\overset{1}{\int }}\log \left(\frac{2-x}{2+x}\right)dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{-1}{\overset{1}{\int }}\log \left(\frac{2-x}{2+x}\right)dx \\ \mathrm{Let}f\left(x\right)=\log \left(\frac{2-x}{2+x}\right) \\ f\left(-\mathrm{x}\right)=\log \left(\frac{2-\left(-x\right)}{2+\left(-x\right)}\right) \\ =\log \left(\frac{2+x}{2-x}\right) \\ =\log {\left(\frac{2-x}{2+x}\right)}^{-1} \\ =-\log \left(\frac{2-x}{2+x}\right) \\ =-f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=0 \end{gathered}$$​

​Hence, the value of the integral $\underset{-1}{\overset{1}{\int }}\log \left(\frac{2-x}{2+x}\right)dx$ is 0.