Answer
$$ - {\operatorname{csch} ^{ - 1}}\left| {2x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {1 + 4{x^2}} }}} \cr
& {\text{substitute }}u = 2x,{\text{ }}du = 2dx \cr
& = \int {\frac{{dx}}{{x\sqrt {1 + 4{x^2}} }}} = \int {\frac{{\left( {1/2} \right)du}}{{\left( {u/2} \right)\sqrt {1 + {u^2}} }}} \cr
& = \int {\frac{{du}}{{u\sqrt {1 + {u^2}} }}} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& = - {\operatorname{csch} ^{ - 1}}\left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = - {\operatorname{csch} ^{ - 1}}\left| {2x} \right| + C \cr} $$
