Answer
$$\ln \left( {2 + \sqrt 3 } \right)\left( {\sqrt 2 - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_{2 + \sqrt 2 }^4 {\frac{{dx}}{{\sqrt {\left( {x - 1} \right)\left( {x - 3} \right)} }}} \cr
& {\text{Use FOIL}} \cr
& = \int_{2 + \sqrt 2 }^4 {\frac{{dx}}{{\sqrt {{x^2} - 3x - x + 3} }}} \cr
& = \int_{2 + \sqrt 2 }^4 {\frac{{dx}}{{\sqrt {{x^2} - 4x + 3} }}} \cr
& {\text{Completing the square}} \cr
& {x^2} - 4x + 3 = {\left( {x - 2} \right)^2} - 1 \cr
& = \int_{2 + \sqrt 2 }^4 {\frac{{dx}}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} \cr
& {\text{Let }}u = x - 2,\,\,\,\,du = dx \cr
& {\text{Change the limits of integration}} \cr
& x = 4,\,\,\,\,u = 2 \cr
& x = 2 + \sqrt 2 ,\,\,\,\,u = \sqrt 2 \cr
& {\text{Use the change of variable}} \cr
& \int_{2 + \sqrt 2 }^4 {\frac{{dx}}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} = \int_{\sqrt 2 }^2 {\frac{{du}}{{\sqrt {{u^2} - 1} }}} \cr
& \cr
& {\text{Let }}u = \sec \theta ,\,\,\,du = \sec \theta \tan \theta d\theta \cr
& \int_{\sqrt 2 }^2 {\frac{{du}}{{\sqrt {{u^2} - 1} }}} = \int_{{{\sec }^{ - 1}}\sqrt 2 }^{{{\sec }^{ - 1}}2} {\frac{{\sec \theta \tan \theta d\theta }}{{\sqrt {{{\sec }^2}\theta - 1} }}} \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{\sec \theta \tan \theta d\theta }}{{\sqrt {{{\tan }^2}\theta } }}} \cr
& = \int_{\pi /4}^{\pi /3} {\sec \theta } d\theta \cr
& \cr
& {\text{Integrate}} \cr
& = \left( {\ln \left| {\sec \theta + \tan \theta } \right|} \right)_{\pi /4}^{\pi /3} \cr
& = \ln \left| {\sec \left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right)} \right| - \ln \left| {\sec \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right)} \right| \cr
& = \ln \left| {2 + \sqrt 3 } \right| + \ln \left| {\sqrt 2 - 1} \right| \cr
& = \ln \left( {2 + \sqrt 3 } \right)\left( {\sqrt 2 - 1} \right) \cr} $$
