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Đặt $\sqrt{x^2+1} = xt \Rightarrow x^2 =\dfrac{1}{t^2-1} \Rightarrow xdx =-\dfrac{t}{(t^2-1)^2}dt$
Ta có $\dfrac{dx}{\sqrt{x^2+1}}=\dfrac{xdx}{x.xt}=\dfrac{-\dfrac{tdt}{(t^2-1)^2}}{\dfrac{1}{t^2-1}t}=-\dfrac{dt}{t^2-1}$
Vậy $I=-\int \limits_{\sqrt 2}^{\frac{\sqrt 5}{2}} \dfrac{1}{\dfrac{1}{t^2-1}. (t^2-1)}dt =-\int\limits_{\sqrt 2}^{\frac{\sqrt 5}{2}} dt$
$=-t \bigg |_{\sqrt 2}^{\frac{\sqrt 5}{2}} =\sqrt 2 -\dfrac{\sqrt 5}{2}$
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