Answer
$$\frac{{\pi \sqrt 2 }}{{48}}$$
Work Step by Step
$$\eqalign{
& \int_{1/2}^{\left( {\sqrt 2 + 3} \right)/\left( {2\sqrt 2 } \right)} {\frac{{dx}}{{8{x^2} - 8x + 11}}} \cr
& \cr
& {\text{Factor the denominator}} \cr
& = \int_{1/2}^{\left( {\sqrt 2 + 3} \right)/\left( {2\sqrt 2 } \right)} {\frac{{dx}}{{8\left( {{x^2} - x + 11/8} \right)}}} \cr
& {\text{Completing the square}} \cr
& = \frac{1}{8}\int_{1/2}^{\left( {\sqrt 2 + 3} \right)/\left( {2\sqrt 2 } \right)} {\frac{{dx}}{{{x^2} - x + 1/4 + 9/8}}} \cr
& = \frac{1}{8}\int_{1/2}^{\left( {\sqrt 2 + 3} \right)/\left( {2\sqrt 2 } \right)} {\frac{{dx}}{{{{\left( {x - 1/2} \right)}^2} + {{\left( {3/2\sqrt 2 } \right)}^2}}}} \cr
& {\text{Let }}u = x - \frac{1}{2},\,\,\,du = dx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{1}{2} \to u = \frac{1}{2} - \frac{1}{2} = 0 \cr
& x = \frac{{\sqrt 2 + 3}}{{2\sqrt 2 }} \to u = \frac{{\sqrt 2 + 3}}{{2\sqrt 2 }} - \frac{1}{2} = \frac{{3\sqrt 2 }}{4} \cr
& \cr
& {\text{Use the change of variable }} \cr
& = \frac{1}{8}\int_0^{\frac{{3\sqrt 2 }}{4}} {\frac{{du}}{{{u^2} + {{\left( {3/2\sqrt 2 } \right)}^2}}}} \cr
& {\text{Integrate}} \cr
& = \frac{1}{8}\left( {\frac{{2\sqrt 2 }}{3}} \right)\left[ {{{\tan }^{ - 1}}\left( {\frac{{2\sqrt 2 u}}{3}} \right)} \right]_0^{\frac{{3\sqrt 2 }}{4}} \cr
& = \frac{{\sqrt 2 }}{{12}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {3\sqrt 2 /4} \right)}}{3}} \right) - 0} \right] \cr
& = \frac{{\sqrt 2 }}{{12}}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - 0} \right] \cr
& = \frac{{\sqrt 2 }}{{12}}\left( {\frac{\pi }{4}} \right) \cr
& = \frac{{\pi \sqrt 2 }}{{48}} \cr} $$
