Cách 2 $$\begin{array}{c}\mathop \smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{dx}}{{\sqrt {{x^2} + 1} }} = \mathop \smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} \left( {x + \sqrt {{x^2} + 1} } \right)}}dx\\ = \mathop \smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{d(x + \sqrt {{x^2} + 1} )}}{{x + \sqrt {{x^2} + 1} }} = \ln |x + \sqrt {{x^2} + 1} |_{\sqrt 3 }^{\sqrt 8 } = \ln |\frac{{3 + 2\sqrt 2 }}{{2 + \sqrt 3 }}|\end{array}$$
Cách 2 $$\smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{dx}}{{\sqrt {{x^2} + 1} }} = \mathop \smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} \left( {x + \sqrt {{x^2} + 1} } \right)}}dx\\ = \mathop \smallint \nolimits_{\sqrt 3 }^{\sqrt 8 } \frac{{d(x + \sqrt {{x^2} + 1} )}}{{x + \sqrt {{x^2} + 1} }} = \ln |x + \sqrt {{x^2} + 1} |_{\sqrt 3 }^{\sqrt 8 } = \ln |\frac{{3 + 2\sqrt 2 }}{{2 + \sqrt 3 }}$$