Answer
$$A = 2{\left( {\ln 5} \right)^2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{4\ln x}}{x},{\text{ }}y = 0,{\text{ }}x = 5 \cr
& {\text{From the graph we can define the area as:}} \cr
& A = \int_1^5 {\frac{{4\ln x}}{x}} dx \cr
& {\text{Integrate}} \cr
& A = 4\left[ {\frac{{{{\left( {\ln x} \right)}^2}}}{2}} \right]_1^5 \cr
& A = 2\left[ {{{\left( {\ln 5} \right)}^2} - {{\left( {\ln 1} \right)}^2}} \right] \cr
& A = 2{\left( {\ln 5} \right)^2} \cr
& A \approx 5.1805 \cr} $$
