Answer
$$\frac{1}{{12}}\pi $$
Work Step by Step
$$\eqalign{
& \int_{1/12}^{1/4} {\frac{{dx}}{{\sqrt x \left( {1 + 4x} \right)}}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int_{1/12}^{1/4} {\frac{{dx}}{{\sqrt x \left( {1 + {{\left( {2\sqrt x } \right)}^2}} \right)}}} \cr
& {\text{Let }}u = 2\sqrt x ,\,\,\,du = \frac{1}{{\sqrt x }}dx \cr
& {\text{Use the substitution}} \cr
& \int_{1/12}^{1/4} {\frac{{dx}}{{\sqrt x \left( {1 + {{\left( {2\sqrt x } \right)}^2}} \right)}}} = \int_{2\sqrt {1/12} }^{2\sqrt {1/4} } {\frac{{du}}{{1 + {u^2}}}} = \int_{\sqrt 3 /3}^1 {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrating}} \cr
& \int_{\sqrt 3 /3}^{1} {\frac{{du}}{{1 + {u^2}}}} = \left[ {{{\tan }^{ - 1}}u} \right]_{\sqrt 3 /3}^{1} \cr
& \left[ {{{\tan }^{ - 1}}u} \right]_{\sqrt 3 /3}^{1} = {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{3}} \right) \cr
& \left[ {{{\tan }^{ - 1}}u} \right]_{\sqrt 3 /3}^{1} = \frac{\pi }{4} - \frac{\pi }{6} \cr
& \left[ {{{\tan }^{ - 1}}u} \right]_{\sqrt 3 /3}^{1} = \frac{1}{{12}}\pi \cr} $$
