Answer
$$ = \frac{1}{4}ta{n^{ - 1}}\left( {4x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + 16{x^2}}}} \cr
& or \cr
& \int {\frac{{dx}}{{1 + {{\left( {4x} \right)}^2}}}} \cr
& {\text{substitutue }}u = 4x,{\text{ }}du = 4dx,{\text{ }}\frac{1}{4}du = dx \cr
& \int {\frac{{dx}}{{1 + {{\left( {4x} \right)}^2}}}} = \int {\frac{{\left( {1/4} \right)du}}{{1 + {u^2}}}} \cr
& = \frac{1}{4}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{4}\left( {\frac{1}{1}ta{n^{ - 1}}\left( {\frac{u}{1}} \right) + C} \right) \cr
& = \frac{1}{4}ta{n^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{4}ta{n^{ - 1}}\left( {4x} \right) + C \cr} $$