Tính tích phân $I = \int\limits_1^{\sqrt 3 } {\dfrac{{\sqrt {1 + {x^2}} }}{{{x^2}}}dx}$ ta được:

Đặt $t = \dfrac{{\sqrt {1 + {x^2}} }}{x}$ $\Leftrightarrow {t^2}{x^2} = 1 + {x^2} \Leftrightarrow {x^2}\left( {{t^2} - 1} \right) = 1 \Rightarrow {x^2} = \dfrac{1}{{{t^2} - 1}}$ $\Rightarrow 2xdx = \dfrac{{ - 2t}}{{{{\left( {{t^2} - 1} \right)}^2}}}dt$

$\Rightarrow \dfrac{{dx}}{x} = \dfrac{{ - tdt}}{{{{\left( {{t^2} - 1} \right)}^2}}}.\left( {{t^2} - 1} \right) = \dfrac{{ - tdt}}{{{t^2} - 1}}$

Đổi cận $\left\{ \begin{array}{l}x = 1 \Rightarrow t = \sqrt 2 \\x = \sqrt 3  \Rightarrow t = \dfrac{2}{{\sqrt 3 }}\end{array} \right.$ , khi đó ta có:

$\begin{array}{l}I =  - \int\limits_{\sqrt 2 }^{\dfrac{2}{{\sqrt 3 }}} {\dfrac{{{t^2}dt}}{{{t^2} - 1}}}  = \int\limits_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 } {\left( {1 + \dfrac{1}{{{t^2} - 1}}} \right)dt}  \\= \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \int\limits_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 } {\dfrac{1}{{{t^2} - 1}}dt}  \\= \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \left. {\dfrac{1}{2}\ln \left| {\dfrac{{t - 1}}{{t + 1}}} \right|} \right|_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 }\\ = \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \dfrac{1}{2}\left( {\ln \left( {3 - 2\sqrt 2 } \right) - \ln \left( {7 - 4\sqrt 3 } \right)} \right)\\ = \sqrt 2  - \dfrac{2}{{\sqrt 3 }} + \ln \left( {\sqrt 2  - 1} \right) + \ln \left( {2 - \sqrt 3 } \right)\\ = \sqrt 2  - \dfrac{2}{{\sqrt 3 }} + \ln \dfrac{{\sqrt 2  - 1}}{{2 - \sqrt 3 }}\end{array}$