Answer
$$2\sqrt x + 2\ln \left| {\frac{{\sqrt x - 2}}{{\sqrt x + 2}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x }}{{x - 4}}} dx \cr
& {\text{Let }}{t^2} = x,{\text{ }}dx = 2tdt \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sqrt x }}{{x - 4}}} dx = \int {\frac{{\sqrt {{t^2}} }}{{{t^2} - 4}}} \left( {2tdt} \right) \cr
& = \int {\frac{{2{t^2}}}{{{t^2} - 4}}} dt \cr
& {\text{By long division }}\frac{{2{t^2}}}{{{t^2} - 4}} = 2 + \frac{8}{{{t^2} - 4}} \cr
& \int {\frac{{2{t^2}}}{{{t^2} - 4}}} d = \int {\left( {2 + \frac{8}{{{t^2} - 4}}} \right)} dt \cr
& = 2t + \int {\frac{8}{{{t^2} - 4}}} dt \cr
& {\text{Decompose }}\frac{8}{{{t^2} - 4}}{\text{ into partial fractions}} \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{8}{{{t^2} - 4}} = \frac{A}{{t - 2}} + \frac{B}{{t + 2}} \cr
& 8 = A\left( {t + 2} \right) + B\left( {t - 2} \right) \cr
& t = 2 \to A = 2 \cr
& t = - 2 \to B = - 2 \cr
& \frac{8}{{{t^2} - 4}} = \frac{2}{{t - 2}} - \frac{2}{{t + 2}} \cr
& = 2t + \int {\frac{8}{{{t^2} - 4}}} dt \cr
& = 2t + \int {\left( {\frac{2}{{t - 2}} - \frac{2}{{t + 2}}} \right)} dt \cr
& {\text{Integrating}} \cr
& = 2t + 2\ln \left| {t - 2} \right| - 2\ln \left| {t + 2} \right| + C \cr
& = 2t + 2\ln \left| {\frac{{t - 2}}{{t + 2}}} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sqrt x {\text{ for }}t \cr
& = 2\sqrt x + 2\ln \left| {\frac{{\sqrt x - 2}}{{\sqrt x + 2}}} \right| + C \cr} $$
