Answer
$$ - \frac{{\sqrt {{x^2} + 1} }}{x} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} + 1} }}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}x = \tan \theta,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} + 1} }}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {\tan \theta } \right)}^2}\sqrt {{{\tan }^2}\theta + 1} }}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta \sqrt {{{\tan }^2}\theta + 1} }}} \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta \sqrt {{{\sec }^2}\theta } }}} \cr
& {\text{simplifying}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta \sec \theta }}} \cr
& = \int {\frac{{\sec \theta d\theta }}{{{{\tan }^2}\theta }}} \cr
& = \int {\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \cr
& = \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}} d\theta \cr
& u = \sin \theta,\,\,\,du = \cos \theta \cr
& = \int {\frac{1}{{{u^2}}}} du \cr
& {\text{Integrate}} \cr
& = - \frac{1}{u} + C \cr
& = - \frac{1}{{\sin \theta }} + C \cr
& {\text{write in terms of }}x,{\text{ }}\sin \theta = \frac{x}{{\sqrt {{x^2} + 1} }} \cr
& = - \frac{{\sqrt {{x^2} + 1} }}{x} + C \cr} $$
