Answer
$$A = \frac{1}{3}\ln \left( {26} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{{{x^3} - 3x}}{\text{ and }}y = \frac{1}{{{x^3} - 3x}}{\text{ on the interval }}\left[ {2,4} \right] \cr
& {\text{From the graph we can see that }}\frac{{{x^2}}}{{{x^3} - 3x}} > \frac{1}{{{x^3} - 3x}}{\text{ on }}\left[ {2,4} \right] \cr
& {\text{The area is given by:}} \cr
& A = \int_2^4 {\left( {\frac{{{x^2}}}{{{x^3} - 3x}} - \frac{1}{{{x^3} - 3x}}} \right)} dx \cr
& A = \int_2^4 {\left( {\frac{{{x^2} - 1}}{{{x^3} - 3x}}} \right)} dx \cr
& A = \frac{1}{3}\int_2^4 {\left( {\frac{{3{x^2} - 3}}{{{x^3} - 3x}}} \right)} dx \cr
& A = \frac{1}{3}\left[ {\ln \left| {{x^3} - 3x} \right|} \right]_2^4 \cr
& A = \frac{1}{3}\left[ {\ln \left| {{{\left( 4 \right)}^3} - 3\left( 4 \right)} \right| - \ln \left| {{{\left( 2 \right)}^3} - 3\left( 2 \right)} \right|} \right] \cr
& A = \frac{1}{3}\left[ {\ln \left| {52} \right| - \ln \left| 2 \right|} \right] \cr
& A = \frac{1}{3}\ln \left( {\frac{{52}}{2}} \right) \cr
& A = \frac{1}{3}\ln \left( {26} \right) \cr} $$
