Answer
$$4 - \pi $$
Work Step by Step
$$\eqalign{
& \int_4^8 {\frac{{\sqrt {x - 4} }}{x}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {u^2} = x - 4,\,\,\,\,\,\,\,u = \sqrt {x - 4} ,\,\,\,\,\,x = {u^2} + 4,\,\,\,\,dx = 2udu \cr
& \,\,\,\,x = 8 \to u = 2\,\,\,\,\,\,\,\,\,\,{\text{ and}}\,\,\,\,\,\,\,\,\,{\text{ }}x = 4,\,\,\,u = 0 \cr
& \cr
& {\text{write the integral in terms of }}u \cr
& \int_4^8 {\frac{{\sqrt {x - 4} }}{x}} dx = \int_0^2 {\frac{{\sqrt {{u^2}} }}{{{u^2} + 4}}} \left( {2udu} \right) \cr
& = \int_0^2 {\frac{{2{u^2}}}{{{u^2} + 4}}} du \cr
& {\text{Use long division}} \cr
& = \int_0^2 {\left( {2 - \frac{8}{{{u^2} + 4}}} \right)} du \cr
& \cr
& {\text{integrate and simplify}} \cr
& = \left[ {2u - \frac{8}{2}{{\tan }^{ - 1}}\left( {\frac{u}{2}} \right)} \right]_0^2 \cr
& = 2\left[ {u - 2{{\tan }^{ - 1}}\left( {\frac{u}{2}} \right)} \right]_0^2 \cr
& = 2\left[ {2 - 2{{\tan }^{ - 1}}\left( {\frac{2}{2}} \right)} \right] - 2\left[ {0 - 2{{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& = 2\left( {2 - 2\left( {\frac{\pi }{4}} \right)} \right) \cr
& = 2\left( {2 - \frac{\pi }{2}} \right) \cr
& = 4 - \pi \cr} $$
