\(\displaystyle \int \frac{8}{x^2-4}  dx \) =

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Integration by partial fractions:

$$ \frac{8}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2} $$ $$ Ax-2A+Bx+2B=8 $$ $$ A+B=0 \hskip{2em} -2A+2B=8 $$ $$ A=-2, B=2 $$ $$ \int \frac{8}{x^2-4}  dx = \int \frac{-2}{x+2} + \frac{2}{x-2}   dx$$ $$ = -2\ln|x+2|+2\ln|x-2|+C $$ $$ = \boxed{2\ln\left|\dfrac{x-2}{x+2} \right| + C } $$