Answer
$$\ln \left| {\sqrt {{{\left( {x - 3} \right)}^2} + 1} + x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {{x^2} - 6x + 10} }}} \cr
& {\text{completing the square for }}{x^2} - 6x + 10 \cr
& {x^2} - 6x + 10 = \left( {{x^2} - 6x + 9} \right) + 1 \cr
& {x^2} - 6x + 10 = {\left( {x - 3} \right)^2} + 1 \cr
& \int {\frac{{dx}}{{\sqrt {{x^2} - 6x + 10} }}} = \int {\frac{{dx}}{{\sqrt {{{\left( {x - 3} \right)}^2} + 1} }}} \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x - 3 = \tan \theta ,{\text{ }}\,\,\,dx = {\sec ^2}\theta d\theta \cr
& \int {\frac{{dx}}{{\sqrt {{{\left( {x - 3} \right)}^2} + 1} }}} = \int {\frac{{{{\sec }^2}\theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}d\theta } \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}d\theta } \cr
& = \int {\sec \theta d\theta } \cr
& {\text{Integrate}} \cr
& = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& \cr
& {\text{Where }}x - 3 = \tan \theta {\text{ and sec}}\theta = \sqrt {{{\left( {x - 3} \right)}^2} + 1} \cr
& = \ln \left| {\sqrt {{{\left( {x - 3} \right)}^2} + 1} + x - 3} \right| + C \cr} $$
