Answer
$$\sqrt 2 - \frac{2}{{\sqrt 3 }} + \ln \left( {\frac{{2 + \sqrt 3 }}{{1 + \sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt 3 } {\frac{{\sqrt {1 + {x^2}} }}{{{x^2}}}} dx \cr
& {\text{Let }}x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{The new limits of integration are:}} \cr
& x = \sqrt 3 \to \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \frac{\pi }{3} \cr
& x = 1 \to \theta = {\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4} \cr
& {\text{Substituting}} \cr
& \int_1^{\sqrt 3 } {\frac{{\sqrt {1 + {x^2}} }}{{{x^2}}}} dx = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {1 + {{\left( {\tan \theta } \right)}^2}} }}{{{{\left( {\tan \theta } \right)}^2}}}\left( {{{\sec }^2}\theta } \right)} d\theta \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {1 + {{\tan }^2}\theta } }}{{{{\tan }^2}\theta }}\left( {{{\sec }^2}\theta } \right)} d\theta \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {{{\sec }^2}\theta } }}{{{{\tan }^2}\theta }}\left( {{{\sec }^2}\theta } \right)} d\theta \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{{{\sec }^2}\theta }}{{{{\tan }^2}\theta }}\left( {\sec \theta } \right)} d\theta \cr
& {\text{Use pythagorean identity}} \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{1 + {{\tan }^2}\theta }}{{{{\tan }^2}\theta }}\left( {\sec \theta } \right)} d\theta \cr
& = \int_{\pi /4}^{\pi /3} {\left( {\frac{{\sec \theta {{\tan }^2}\theta }}{{{{\tan }^2}\theta }} + \frac{{\sec \theta }}{{{{\tan }^2}\theta }}} \right)} d\theta \cr
& = \int_{\pi /4}^{\pi /3} {\left( {\sec \theta + \csc \theta \cot \theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \left[ {\ln \left| {\sec \theta + \tan \theta } \right| - \csc \theta } \right]_{\pi /4}^{\pi /3} \cr
& {\text{Evaluating}} \cr
& = \left[ {\ln \left| {\sec \frac{\pi }{3} + \tan \frac{\pi }{3}} \right| - \csc \frac{\pi }{3}} \right] - \left[ {\ln \left| {\sec \frac{\pi }{4} + \tan \frac{\pi }{4}} \right| - \csc \frac{\pi }{4}} \right] \cr
& = \left[ {\ln \left| {2 + \sqrt 3 } \right| - \frac{{2\sqrt 3 }}{3}} \right] - \left[ {\ln \left| {\sqrt 2 + 1} \right| - \sqrt 2 } \right] \cr
& = \sqrt 2 - \frac{2}{{\sqrt 3 }} + \ln \left( {2 + \sqrt 3 } \right) - \ln \left( {1 + \sqrt 2 } \right) \cr
& = \sqrt 2 - \frac{2}{{\sqrt 3 }} + \ln \left( {\frac{{2 + \sqrt 3 }}{{1 + \sqrt 2 }}} \right) \cr} $$
