$\underset{-1}{\overset{1}{\int }}{\{{\left(\frac{x+2}{x-2}\right)}^2+{\left(\frac{x-2}{x+2}\right)}^2-2\}}^{\frac{1}{2}}dx=$

$I=\underset{-1}{\overset{1}{\int }}{\{{\left(\frac{x+2}{x-2}\right)}^2+{\left(\frac{x-2}{x+2}\right)}^2-2\}}^{\frac{1}{2}}dx=\underset{-1}{\overset{1}{\int }}{\left\{{(\frac{x+2}{x-2}-\frac{x-2}{x+2})}^2\right\}}^{\frac{1}{2}}dx$

$=2\underset{-1}{\overset{1}{\int }}\left|\frac{8x}{x^2-4}\right|dx$

$=-2\times 4\underset{0}{\overset{1}{\int }}\frac{-2x}{4-x^2}dx$

$=-8\left[\log {\left[|4-x^2|\right]}_0^1\right]$

$=-8[\log \left|3\right|-\log \left|4\right|]$

$=-8\log \left|\frac{3}{4}\right|$

$=8\log \left|\frac{{\displaystyle 4}}{{\displaystyle 3}}\right|$