Answer
$$\frac{1}{2}{\tan ^{ - 1}}2x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + 4{x^2}}}} \cr
& {\text{substitute }}u = 2x,{\text{ }}du = 2dx{\text{ and }}dx = \frac{{du}}{2} \cr
& \int {\frac{{dx}}{{1 + 4{x^2}}}} = \int {\frac{{du/2}}{{1 + {u^2}}}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}{\tan ^{ - 1}}u + C \cr
& {\text{ with}}\,\,\,u = 2x \cr
& = \frac{1}{2}{\tan ^{ - 1}}2x + C \cr} $$
