Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_{1/12}^{1/4} {\frac{{2dt}}{{\sqrt t + 4t\sqrt t }}} \cr
& {\text{Factoring the denominator}} \cr
& = \int_{1/12}^{1/4} {\frac{{2dt}}{{\sqrt t \left( {1 + 4t} \right)}}} \cr
& {\text{Write as}} \cr
& = \int_{1/12}^{1/4} {\frac{{2dt}}{{\sqrt t \left( {1 + {{\left( {2\sqrt t } \right)}^2}} \right)}}} \cr
& {\text{We set}}{\text{, }}2\sqrt t = \tan \theta,{\text{ }}\frac{1}{{\sqrt t }}dt = {\sec ^2}\theta d\theta \cr
& \theta = {\tan ^{ - 1}}\left( {{\text{ }}2\sqrt t } \right) \cr
& {\text{The new limits on }}\theta {\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 1/4,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {{\text{ }}2\sqrt {1/4} } \right) = \pi /4 \cr
& \,\,\,\,\,\,{\text{If }}t = 1/2,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {{\text{ }}2\sqrt {1/12} } \right) = \pi /6 \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int_{1/12}^{1/4} {\frac{{2dt}}{{\sqrt t \left( {1 + {{\left( {2\sqrt t } \right)}^2}} \right)}}} = \int_{\pi /6}^{\pi /4} {\frac{{2{{\sec }^2}\theta d\theta }}{{1 + {{\tan }^2}\theta }}} \cr
& = 2\int_{\pi /6}^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{1 + {{\tan }^2}\theta }}} \cr
& {\text{use the fundamental identity }}{\tan ^2}x + 1 = {\sec ^2}x \cr
& = 2\int_{\pi /6}^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^2}\theta }}} \cr
& = 2\int_{\pi /6}^{\pi /4} {\theta d\theta } \cr
& {\text{integrating}} \cr
& = 2\left( \theta \right)_{\pi /6}^{\pi /4} \cr
& = 2\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr
& = \frac{\pi }{6} \cr} $$
