Answer
$$area = \frac{{16}}{3}\ln \left( 4 \right) - 4$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \sqrt x \ln x,\,\,\,\,y = 0,\,\,\,x = 4 \cr
& {\text{For }}y = 0\,,\,\,\,\,\sqrt x \ln x = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \cr
& \cr
& {\text{We can represents the enclosed by the curves as}} \cr
& {\text{area}} = \int_1^4 {\sqrt x \ln x} dx \cr
& {\text{integrate by parts}} \cr
& \,\,\,\,\,\,{\text{let }}u = \ln x \to du = \frac{1}{x}dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,dv = {x^{1/2}}dx \to v = \frac{2}{3}{x^{3/2}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\sqrt x \ln x} = {x^{3/2}}\left( {\frac{2}{3}\ln x} \right) - \int {\left( {\frac{2}{3}{x^{3/2}}} \right)} \left( {\frac{1}{x}} \right)dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{3}{x^{3/2}}\ln x - \frac{2}{3}\int {{x^{1/2}}} dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{3}{x^{3/2}}\ln x - \frac{2}{3}\left( {\frac{2}{3}{x^{3/2}}} \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{3}{x^{3/2}}\ln x - \frac{4}{9}{x^{3/2}} + C \cr
& \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \left[ {\frac{2}{3}{x^{3/2}}\ln x - \frac{4}{9}{x^{3/2}}} \right]_1^4 \cr
& {\text{evaluate the limits and simplify}} \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}}\ln \left( 4 \right) - \frac{4}{9}{{\left( 4 \right)}^{3/2}}} \right] - \left[ {\frac{2}{3}{{\left( 1 \right)}^{3/2}}\ln \left( 1 \right) - \frac{4}{9}{{\left( 1 \right)}^{3/2}}} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}}\ln \left( 4 \right) - \frac{4}{9}{{\left( 4 \right)}^{3/2}}} \right] - \left[ {\frac{2}{3}{{\left( 1 \right)}^{3/2}}\ln \left( 1 \right) - \frac{4}{9}{{\left( 1 \right)}^{3/2}}} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \left[ {\frac{{16}}{3}\ln \left( 4 \right) - \frac{{32}}{9}} \right] - \left[ {\frac{4}{9}} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \frac{{16}}{3}\ln \left( 4 \right) - \frac{{32}}{9} - \frac{4}{9} \cr
& \,\,\,\,\,\,\,\,\,\,\,area = \frac{{16}}{3}\ln \left( 4 \right) - 4 \cr} $$
