Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {{x^2} - 1} = t \Rightarrow {x^2} - 1 = {t^2}\\
\Rightarrow \left\{ \begin{array}{l}
dx = \dfrac{{tdt}}{x}\\
{x^2} = {t^2} + 1
\end{array} \right.\\
x = \sqrt 2 \Rightarrow t = 1,x = 2 \Rightarrow t = \sqrt 3 \\
I = \dfrac{1}{2}\int\limits_1^{\sqrt 3 } {\dfrac{{{t^2}}}{{{t^2} + 1}}dt} = \dfrac{1}{2}\int\limits_1^{\sqrt 3 } {\left( {1 - \dfrac{1}{{{t^2} + 1}}} \right)dt} \\
= \dfrac{1}{2}\left. t \right|_1^{\sqrt 3 } - \dfrac{1}{2}\int\limits_1^{\sqrt 3 } {\dfrac{1}{{{t^2} + 1}}} dt = \dfrac{{\sqrt 3 - 1}}{2} - \dfrac{1}{2}{I_2}\\
{I_2} = \int\limits_1^{\sqrt 3 } {\dfrac{1}{{{t^2} + 1}}} dt\\
t = \tan u \Rightarrow dt = \left( {1 + {t^2}} \right)du\\
\Rightarrow {I_2} = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{1}{{{t^2} + 1}}.\left( {{t^2} + 1} \right)du} = \dfrac{\pi }{3} - \dfrac{\pi }{4} = \dfrac{\pi }{{12}}\\
I = \dfrac{{\sqrt 3 - 1}}{2} - \dfrac{\pi }{{24}}
\end{array}\)
