Answer
$$\frac{x}{{\sqrt {{x^2} + 1} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}} \cr
& {\text{The integrand contains the form }}{a^2} + {x^2} \cr
& 1 + {x^2} \to a = 1 \cr
& {\text{Use the change of variable }}x = a\tan \theta \cr
& x = \tan \theta ,\,\,\,dx = {\sec ^2}\theta d\theta \cr
& \cr
& {\text{Substituting}} \cr
& \int {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}}}} \cr
& {\text{Simplifying}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^3}\theta }}} \cr
& = \int {\frac{1}{{\sec \theta }}} d\theta \cr
& = \int {\cos \theta } d\theta \cr
& {\text{Integrating}} \cr
& = \sin \theta + C \cr
& \cr
& {\text{Write in terms of }}x \cr
& \tan \theta = x \to \frac{{opp}}{{ady}} = \frac{x}{1},\,\,\,hyp = \sqrt {{x^2} + 1} \cr
& \sin \theta = \frac{{opp}}{{hyp}} = \frac{x}{{\sqrt {{x^2} + 1} }} \cr
& ,{\text{then}} \cr
& = \frac{x}{{\sqrt {{x^2} + 1} }} + C \cr} $$
