The integral  ${\int }_{-1/2}^{1/2} \left(\left[x\right]+2\log \frac{1+x}{1-x}\right)\mathrm{d}x$ equal  ((where [x] is greatest integer function)

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

(since log $\frac{1+x}{1-x}$ is an odd function)

$={\int }_{-1/2}^0 \left[x\right]dx+{\int }_0^{1/2} \left[x\right]dx={\int }_{-1/2}^0 (-1)dx=-\frac{1}{2}.$