Answer
$$Area = 2 + \frac{4}{9}\ln \left( {\frac{8}{{17}}} \right) + \ln \left( {\frac{2}{5}} \right)$$
Work Step by Step
$$\eqalign{
& \frac{1}{x} = \frac{x}{{3x + 4}} \cr
& {\text{solve for }}x \cr
& 3x + 4 = {x^2} \cr
& {x^2} - 3x - 4 = 0 \cr
& \left( {x - 4} \right)\left( {x + 1} \right) = 0 \cr
& x = 4,{\text{ }}x = - 1,{\text{ and the line }}x = 10 \cr
& {\text{the interval is }}\left[ {4,10} \right] \cr
& \frac{x}{{3x + 4}} \geqslant \frac{1}{x}{\text{ in the interval }}\left[ {4,10} \right] \cr
& {\text{the area of the region is}} \cr
& Area = \int_4^{10} {\left( {\frac{x}{{3x + 4}} - \frac{1}{x}} \right)} dx \cr
& {\text{use long division for }}\frac{x}{{3x + 4}} \cr
& Area = \int_4^{10} {\left( {\frac{1}{3} - \frac{{4/3}}{{3x + 4}} - \frac{1}{x}} \right)} dx \cr
& {\text{integrating}} \cr
& Area = \left( {\frac{1}{3}x - \frac{4}{9}\ln \left| {3x + 4} \right| - \ln \left| x \right|} \right)_4^{10} \cr
& {\text{evaluate limits}} \cr
& Area = \left( {\frac{1}{3}\left( {10} \right) - \frac{4}{9}\ln \left| {3\left( {10} \right) + 4} \right| - \ln \left| {10} \right|} \right) - \left( {\frac{1}{3}\left( 4 \right) - \frac{4}{9}\ln \left| {3\left( 4 \right) + 4} \right| - \ln \left| 4 \right|} \right) \cr
& Area = \frac{{10}}{3} - \frac{4}{9}\ln \left| {\left( {34} \right)} \right| - \ln \left( {10} \right) - \frac{4}{3} + \frac{4}{9}\ln \left( {16} \right) + \ln \left( 4 \right) \cr
& {\text{properties of logarithms}} \cr
& Area = 2 + \frac{4}{9}\ln \left( {\frac{8}{{17}}} \right) + \ln \left( {\frac{2}{5}} \right) \cr} $$
