Biến đổi:
$\int\limits f(x)dx=\int\limits \frac{1-\frac{1}{x^2} }{\frac{1}{x^2}+x^2 } dx=\int\limits \frac{1-\frac{1}{x^2} }{(x+\frac{1}{x})^2-2 } $
Đặt $t=x+\frac{1}{x} $, suy ra:
$dt=(1-\frac{1}{x^2})dx ; \frac{ \left ( 1-\frac{1}{x^2} \right )dx }{(x+\frac{1}{x})^2-2 }=\frac{dt}{t^2-2} $
Khi đó:
$\int\limits f(x)dx=\int\limits \frac{dt}{t^2-2} = \int\limits \frac{1}{2 \sqrt{2} }\left[ {\frac{1}{t- \sqrt{2} }- \frac{1}{t + \sqrt{2} } } \right] dt$
$=\frac{1}{2 \sqrt{2} }\ln|\frac{t-\sqrt{2} }{t+\sqrt{2} } |+C= \frac{1}{2 \sqrt{2} }\ln|\frac{x+\frac{1}{x} -\sqrt{2} }{x+\frac{1}{x} +\sqrt{2} } |+C$
$=\frac{1}{2 \sqrt{2} }\ln|\frac{x^2-x\sqrt{2}+1 }{x^2+x\sqrt{2}+1 } |+C$