Answer
$$A = 3\ln 10$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{6x}}{{1 + {x^2}}} \cr
& f\left( x \right) \geqslant 0{\text{ on the interval }}0 \leqslant x \leqslant 3 \cr
& {\text{From the graph shown below, we can define the area enclosed}} \cr
& {\text{as:}} \cr
& A = \int_0^3 {\frac{{6x}}{{1 + {x^2}}}} dx \cr
& A = 3\int_0^3 {\frac{{2x}}{{1 + {x^2}}}} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = 3\left[ {\ln \left( {1 + {x^2}} \right)} \right]_0^3 \cr
& A = 3\left[ {\ln \left( {1 + {{\left( 3 \right)}^2}} \right) - \ln \left( {1 + {{\left( 0 \right)}^2}} \right)} \right] \cr
& A = 3\left[ {\ln \left( {10} \right) - \ln \left( 1 \right)} \right] \cr
& A = 3\ln 10 \cr} $$
