Ta có

$I=\int\limits_{1/2}^{1}{\frac{{{x}^{2}}-1}{{{x}^{4}}+1}dx}=\int\limits_{1/2}^{1}{\frac{1-\frac{1}{{{x}^{2}}}}{{{x}^{2}}+\frac{1}{{{x}^{2}}}}dx}$

$\xrightarrow{t=x+\frac{1}{x}}I=\int\limits_{5/2}^{2}{\frac{du}{{{u}^{2}}-2}=\left. \left( -\frac{1}{2\sqrt{2}}\ln \left| \frac{\sqrt{2}-u}{\sqrt{2}+u} \right| \right) \right|}_{2}^{5/2}$

$I=-\frac{1}{2\sqrt{2}}\ln \left( \frac{\sqrt{2}-5/2}{\sqrt{2}+5/2}.\frac{\sqrt{2}+2}{\sqrt{2}-2} \right)=\frac{1}{2\sqrt{2}}\ln \left( \frac{6-\sqrt{2}}{6+\sqrt{2}} \right)$

Đáp án đúng là câu A