$\int\limits_{ - 1}^1 {{x^2}\ln \frac{{x + 2}}{{x - 2}}dx}  = \left| {\begin{array}{*{20}{c}}
   {u = \ln \frac{{x + 2}}{{x - 2}}} & {dv = {x^2}}  \\
   {du =  - \frac{4}{{{x^2} - 4}}} & {v = \frac{{{x^3}}}{3}}  \\
\end{array}} \right| = \left. {\frac{{{x^3}}}{3}\ln \frac{{x + 2}}{{x - 2}}} \right|_{ - 1}^1 + \frac{4}{3}\int\limits_{ - 1}^1 {\frac{{{x^3}}}{{{x^2} - 4}}dx}  = \left. {\frac{{{x^3}}}{3}\ln \frac{{x + 2}}{{x - 2}}} \right|_{ - 1}^1 + \frac{4}{3}\int\limits_{ - 1}^1 {\left( {x + \frac{{4x}}{{{x^2} - 4}}} \right)dx}  = \left. {\frac{{{x^3}}}{3}\ln \frac{{x + 2}}{{x - 2}}} \right|_{ - 1}^1 + \left. {\frac{4}{3}\frac{{{x^2}}}{2}} \right|_{ - 1}^1 + \frac{{16}}{6}\int\limits_{ - 1}^1 {\frac{{2x}}{{{x^2} - 4}}dx}  = \left| \begin{array}{l}
 {x^2} - 4 = t \\ 
 2xdx = dt \\ 
 \end{array} \right| = ...$