Answer
$$ - \ln \left| {\frac{{\sqrt {1 + {t^2}} + 1}}{t}} \right| + \sqrt {1 + {t^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 + {t^2}} }}{t}} dt \cr
& {\text{substitute }}t = \tan \theta ,{\text{ }}dt = se{c^2}\theta d\theta \cr
& = \int {\frac{{\sqrt {1 + {{\left( {\tan \theta } \right)}^2}} }}{{\tan \theta }}} se{c^2}\theta d\theta \cr
& {\text{simplify}} \cr
& = \int {\frac{{\sqrt {1 + {{\tan }^2}\theta } }}{{\tan \theta }}} se{c^2}\theta d\theta \cr
& = \int {\frac{{\sec \theta }}{{\tan \theta }}} se{c^2}\theta d\theta \cr
& = \int {\frac{{{{\sec }^3}\theta }}{{\tan \theta }}} d\theta \cr
& = \int {\frac{{{{\sec }^3}\theta }}{{\tan \theta }}} d\theta = \int {\frac{{{{\sec }^2}\theta \sec \theta }}{{\tan \theta }}} d\theta \cr
& {\text{identity se}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta \cr
& = \int {\frac{{\left( {1 + {{\tan }^2}\theta } \right)\sec \theta }}{{\tan \theta }}} d\theta \cr
& {\text{distribute and simplify}} \cr
& = \int {\frac{{\sec \theta + \sec \theta ta{n^2}\theta }}{{\tan \theta }}} d\theta \cr
& = \int {\left( {\frac{{\sec \theta }}{{\tan \theta }} + \frac{{\sec \theta ta{n^2}\theta }}{{\tan \theta }}} \right)} d\theta \cr
& = \int {\left( {\frac{1}{{\sin \theta }} + \sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\left( {\csc \theta + \sec \theta \tan \theta } \right)} d\theta \cr
& {\text{find antiderivatives}} \cr
& = - \ln \left| {\csc \theta + \cot \theta } \right| + \sec \theta + C \cr
& {\text{write in terms of }}t \cr
& = - \ln \left| {\frac{{\sqrt {1 + {t^2}} }}{t} + \frac{1}{t}} \right| + \sqrt {1 + {t^2}} + C \cr
& = - \ln \left| {\frac{{\sqrt {1 + {t^2}} + 1}}{t}} \right| + \sqrt {1 + {t^2}} + C \cr} $$
