\[\begin{array}{l}
I = \int\limits_1^{\sqrt 3 } {\dfrac{{\sqrt {{x^2} + 1} }}{{{x^2}}}dx} \\
x = \tan t \Rightarrow dx = \dfrac{1}{{{{\cos }^2}t}}dt\\
x = 1 \Rightarrow t = \dfrac{\pi }{4}\\
x = \sqrt 3 \Rightarrow t = \dfrac{\pi }{3}\\
I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {{{\tan }^2}t + 1} }}{{{{\tan }^2}t{{\cos }^2}t}}dt} = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{{\dfrac{1}{{\cos t}}}}{{{{\sin }^2}t}}dt} = \dfrac{1}{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{1}{{\sin 2t\sin t}}dt} = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{1}{{\cos t - \cos 3t}}dt}
\end{array}\]
em cũng làm giống chị nhưng tới đây thì thua rồi@@