Answer
$$A = \ln 4$$
Work Step by Step
$$\eqalign{
& {\text{Let the graphs }}y = {x^{ - 1}},\,\,\,y = 4x,\,\,\,\,y = \frac{x}{4} \cr
& {\text{The enclosed area is given by}} \cr
& A = \int_0^{1/2} {\left( {4x - \frac{x}{4}} \right)} dx + \int_{1/2}^2 {\left( {\frac{1}{x} - \frac{x}{4}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {2{x^2} - \frac{{{x^2}}}{8}} \right]_0^{1/2} + \left[ {\ln \left| x \right| - \frac{{{x^2}}}{8}} \right]_{1/2}^2 \cr
& A = \left[ {2{{\left( {\frac{1}{2}} \right)}^2} - \frac{{{{\left( {1/2} \right)}^2}}}{8}} \right] - \left[ 0 \right] + \left[ {\ln \left| 2 \right| - \frac{{{{\left( 2 \right)}^2}}}{8}} \right] - \left[ {\ln \left| {\frac{1}{2}} \right| - \frac{{{{\left( {1/2} \right)}^2}}}{8}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{15}}{{32}} - 0 + \ln 2 - \frac{1}{2} - \ln \frac{1}{2} + \frac{1}{{32}} \cr
& A = 2\ln 2 \cr
& A = \ln 4 \cr} $$
