The integral \(\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _c\left(\frac{1+x}{1-x}\right)\right) \mathrm{d} x\), where \([x]\) represent greatest integer function, equals

Let \(\mathrm{I}=\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right)\right) \mathrm{d} x\)
\(=\int_{\frac{-1}{2}}^0[x] \mathrm{d} x+\int_0^{\frac{1}{2}}[x] \mathrm{d} x+\int_{\frac{-1}{2}}^{\frac{1}{2}} \log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right) \mathrm{d} x\)
Let \(g(x)=\log \left(\frac{1+x}{1-x}\right)\)
\(g(-x)=\log \left(\frac{1-x}{1+x}\right)=-\log \left(\frac{1+x}{1-x}\right)=-g(x)\)
\(\therefore \quad \mathrm{g}(x)\) is a odd function.
\(\therefore \quad \int_{\frac{-1}{2}}^{\frac{1}{2}} \mathrm{~g}(x) \mathrm{d} x=0\)
\(\begin{aligned} \therefore \quad \mathrm{I} & =\int_{\frac{-1}{2}}^0(-1) \mathrm{d} x+\int_0^{\frac{1}{2}}(0) \mathrm{d} x+0 \\ & =[-x]_{\frac{-1}{2}}^0+0 \\ & =\frac{-1}{2}\end{aligned}\)