Answer
$$\ln \left( {\frac{{3 + \sqrt 8 }}{{2 + \sqrt 3 }}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{1/6}^{1/4} {\frac{{dt}}{{t\sqrt {1 - 4{t^2}} }}} \cr
& {\text{set }}u = 2t{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = 2dt,\,\,\,\,\,\,dt = \frac{1}{2}du \cr
& {\text{switch the limits of integration}} \cr
& u = 2t,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = 1/6 \to u = 1/3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = 1/4 \to u = 1/2 \cr
& {\text{use the change of variable}} \cr
& = \int_{1/3}^{1/2} {\frac{{\left( {1/2} \right)du}}{{\left( {u/2} \right)\sqrt {1 - {u^2}} }}} \cr
& = \int_{1/3}^{1/2} {\frac{{du}}{{u\sqrt {1 - {u^2}} }}} \cr
& {\text{Using the Integral formulas in the Theorem 6}}{\text{.12}}:{\text{ }} \cr
& \int {\frac{{dx}}{{x\sqrt {{a^2} - {x^2}} }} = - \frac{1}{a}{{\operatorname{sech} }^{ - 1}}\frac{x}{a} + C,{\text{ for }}0 < x < a} {\text{ then}}{\text{,}} \cr
& = - \left( {{{\operatorname{sech} }^{ - 1}}u} \right)_{1/3}^{1/2} \cr
& {\text{evaluate the limits }} \cr
& = - \left( {{{\operatorname{sech} }^{ - 1}}\left( {\frac{1}{2}} \right) - {{\operatorname{sech} }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) \cr
& = {\operatorname{sech} ^{ - 1}}\left( {\frac{1}{3}} \right) - {\operatorname{sech} ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}:{\text{ }} \cr
& {\operatorname{sech} ^{ - 1}}x = {\cosh ^{ - 1}}\left( {\frac{1}{x}} \right){\text{ for }}\left( {0 < x \leqslant 1} \right).{\text{ Then}}{\text{,}} \cr
& = {\cosh ^{ - 1}}\left( 3 \right) - {\cosh ^{ - 1}}\left( 2 \right) \cr
& {\text{And from the theorem 6}}{\text{.10 we have that }} \cr
& {\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right){\text{ }}\left( {x \geqslant 1} \right){\text{ then}}{\text{,}} \cr
& = \ln \left( {3 + \sqrt {{3^2} - 1} } \right) - \ln \left( {2 + \sqrt {{2^2} - 1} } \right) \cr
& = \ln \left( {3 + \sqrt 8 } \right) - \ln \left( {2 + \sqrt 3 } \right) \cr
& = \ln \left( {\frac{{3 + \sqrt 8 }}{{2 + \sqrt 3 }}} \right) \cr} $$
