#### Answer

$$\ln \left| x \right| + 2{x^{1/2}} + C$$

#### Work Step by Step

$$\eqalign{
& \int {\frac{{1 + \sqrt x }}{x}} dx \cr
& {\text{split the integrand}} \cr
& \int {\left( {\frac{1}{x} + \frac{{\sqrt x }}{x}} \right)} dx \cr
& \int {\left( {\frac{1}{x} + {x^{ - 1/2}}} \right)} dx \cr
& {\text{integrate}} \cr
& = \ln \left| x \right| + \frac{{{x^{1/2}}}}{{1/2}} + C \cr
& = \ln \left| x \right| + 2{x^{1/2}} + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( {\ln \left| x \right| + 2{x^{1/2}} + C} \right) \cr
& {\text{ = }}\frac{d}{{dx}}\left( {\ln \left| x \right|} \right) + \frac{d}{{dx}}\left( {2{x^{1/2}}} \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{ = }}\frac{1}{x} + 2\left( {\frac{1}{2}} \right){x^{ - 1/2}} + 0 \cr
& {\text{simplify}} \cr
& {\text{ = }}\frac{1}{x} + \frac{1}{{\sqrt x }} \cr
& add \cr
& = \frac{{1 + \sqrt x }}{x} \cr} $$