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$\int\limits_{2}^{+\infty }\left (\frac{1}{x^2-1}+\frac{2}{(x+1)^2} \right )dx=\int\limits_{2}^{+\infty }\left (\frac{1}{2(x-1)}-\frac{1}{2(x+1)}+\frac{2}{(x+1)^2} \right )dx$
$=\left[ {\frac{1}{2}\ln|x-1|-\frac{1}{2}\ln|x+1|-\frac{2}{x+1}} \right]_{2}^{+\infty }=\left[ {\frac{1}{2}\ln\dfrac{|x-1|}{|x+1}-\frac{2}{x+1}} \right]_{2}^{+\infty }=\dfrac{2}{3}+\dfrac{\ln 3}{2}$
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