Answer
$$\frac{x}{{\sqrt {4{x^2} + 1} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {1 + 4{x^2}} \right)}^{3/2}}}}} \cr
& {\text{Let }}u = 2x,\,\,\,du = 2dx \cr
& = \frac{1}{2}\int {\frac{{du}}{{{{\left( {1 + {u^2}} \right)}^{3/2}}}}} \cr
& {\text{The integrand contains the form }}{a^2} + {u^2} \cr
& 1 + {u^2} \to a = 1 \cr
& {\text{Use the change of variable }}u = a\tan \theta \cr
& u = \tan \theta ,\,\,\,du = {\sec ^2}\theta d\theta \cr
& \cr
& {\text{Substituting}} \cr
& \frac{1}{2}\int {\frac{{du}}{{{{\left( {1 + {u^2}} \right)}^{3/2}}}}} = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}}}} \cr
& = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} \cr
& = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta d\theta }}{{\sec \theta }}} \cr
& = \frac{1}{2}\int {\frac{1}{{\sec \theta }}} d\theta \cr
& = \frac{1}{2}\int {\cos \theta } d\theta \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{2}\sin \theta + C \cr
& \cr
& {\text{Write in terms of }}u \cr
& {\text{ = }}\frac{1}{2}\left( {\frac{u}{{\sqrt {{u^2} + 1} }}} \right) + C \cr
& {\text{Write in terms of }}x \cr
& {\text{ = }}\frac{1}{2}\left( {\frac{{2x}}{{\sqrt {4{x^2} + 1} }}} \right) + C \cr
& {\text{ = }}\frac{x}{{\sqrt {4{x^2} + 1} }} + C \cr} $$
