${\int }_{-1/2}^{1/2} \left\{\left[x\right]+\log \left(\frac{1+x}{1-x}\right)\right\}dx=$

Given that

${\int }_{-1/2}^{1/2} \left\{\left[x\right]+\log \left(\frac{1+x}{1-x}\right)\right\}dx$

Let $f\left(x\right) = \log \left(\frac{1+x}{1-x}\right) \Rightarrow f\left(-x\right) = \log \left(\frac{1-x}{1+x}\right)$

$f\left(-x\right) = \log {\left(\frac{1+x}{1-x}\right)}^{-1}\Rightarrow f\left(-x\right) = -\log \left(\frac{1+x}{1-x}\right)$

$\Rightarrow f\left(-x\right) =-f\left(x\right)$

So $f\left(x\right)$ is an odd function $\Rightarrow {\int }_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}f\left(x\right)dx = 0 ; f\left(-x\right) =-f\left(x\right)$

${\int }_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[x\right]dx = {\int }_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\left[x\right]dx + {\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[x\right]dx$

$= {\int }_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0-1dx + {\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}0dx$

$={\left[-x\right]}_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0$

$=\frac{-1}{2}.$

$\therefore {\int }_{-1/2}^{1/2} \left\{\left[x\right]+\log \left(\frac{1+x}{1-x}\right)\right\}dx = \frac{-1}{2}.$