The integral ${\int }_{-1/2}^{1/2}\left(\left[x\right]+\ln \left(\frac{1+x}{1-x}\right)\right)dx$equals where [ ] denotes greatest integer function 

given $I=$${\int }_{-1/2}^{1/2}\left(\left[x\right]+\ln \left(\frac{1+x}{1-x}\right)\right)dx$
[ ] denotes greatest integer function

$$\begin{gathered} I={\int }_{-\frac{1}{2}}^{\frac{1}{2}} \left[x\right]dx +{\int }_{-\frac{1}{2}}^{\frac{1}{2}} \ln \left(\frac{1+x}{1-x}\right)dx \\ let I=I_1+I_2 \end{gathered}$$

graph of $y=\left[x\right]$:

$\left[x\right]=\left\{\begin{array}{l}0 , x\in [0,\frac{-1}{2}]\\ -1, x\in [\frac{-1}{2},0]\end{array}\right.$

from graph we get integral =area of shaded region 

$$\begin{gathered} {\int }_{-1/2}^{1/2}\left[x\right] dx ={\int }_{-1/2}^0(-1) dx +{\int }_0^{1/2}0 dx \\ \Rightarrow I_1={\int }_{-1/2}^0-x =-(0-(-\frac{1}{2}\left)\right)=-\frac{1}{2} \\ {I}_2={\int }_{-1/2}^{1/2}\ln \left(\frac{1+x}{1-x}\right)dx \end{gathered}$$

by kings rule  which states that for $$\begin{gathered} I={\int }_a^{l}f\left(x\right) dx \\ {\int }_a^{l}f(a+l-x)dx ={\int }_a^{l}f\left(x\right) dx \\ for I_2={\int }_{-1/2}^{1/2}\ln \left(\frac{1+x}{1-x}\right)dx , a=\frac{1}{2}, l=-\frac{1}{2}, f\left(x\right)=\ln \left(\frac{1+x}{1-x}\right) \\ f(a+l-x)=f(\frac{1}{2}-\frac{1}{2}-x)=f(-x)=\ln \left(\frac{1+x}{1-x}\right)=-\ln \left(\frac{1+x}{1-x}\right) \end{gathered}$$

adding above $I_0$to original $I_2$ we get 

$$\begin{gathered} I_2+I_1=0 \\ \Rightarrow I_2=0 \\ I=I_1+I_2=-\frac{1}{2}+0=-\frac{1}{2} \\ I=-\frac{1}{2} \end{gathered}$$