Ta có:
$I=\int\limits_1^2\dfrac{x^2+1}{(x^2+2x-1)(x^2-2x-1)}dx$
$=\dfrac{1}{2}\int\limits_1^2\left(\dfrac{x-1}{x^2-2x-1}-\dfrac{x+1}{x^2+2x-1}\right)dx$
$=\dfrac{1}{4}\int\limits_1^2\left(\dfrac{d(x^2-2x-1)}{x^2-2x-1}-\dfrac{d(x^2+2x-1)}{x^2+2x-1}\right)dx$
$=\dfrac{1}{4}\ln\dfrac{|x^2-2x-1|}{|x^2+2x-1|}\left|\begin{array}{l}2\\1\end{array}\right.=\dfrac{-\ln7}{4}$