Answer
$$\frac{{x - 2}}{2}\sqrt {{x^2} - 4x + 8} + 2\ln \left( {x - 2 + \sqrt {{x^2} - 4x + 8} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {{x^2} - 4x + 8} dx} \cr
& {\text{completing the square}} \cr
& = \int {\sqrt {{x^2} - 4x + 4 + 4} dx} = \int {\sqrt {\left( {{x^2} - 4x + 4} \right) + 4} dx} \cr
& = \int {\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( 2 \right)}^2}} dx} \cr
& {\text{substitute }}u = x - 2,{\text{ }}du = dx,{\text{ }}a = 2 \cr
& = \int {\sqrt {{u^2} + {a^2}} du} \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{reduction formula 14}} \cr
& = \int {\sqrt {{u^2} + {a^2}} du} = \frac{u}{2}\sqrt {{a^2} + {u^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{a^2} + {u^2}} } \right) + C \cr
& u = x - 2,{\text{ }}a = 2 \cr
& = \frac{{x - 2}}{2}\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( 2 \right)}^2}} + \frac{4}{2}\ln \left( {x - 2 + \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( 2 \right)}^2}} } \right) + C \cr
& = \frac{{x - 2}}{2}\sqrt {{x^2} - 4x + 8} + 2\ln \left( {x - 2 + \sqrt {{x^2} - 4x + 8} } \right) + C \cr} $$
