Answer
${f_{avg}} = \frac{1}{{24}}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2}}}{{{{\left( {{x^3} + 3} \right)}^2}}},{\text{ }}\left[ { - 1,1} \right] \cr
& {\text{Use The Mean Value Theorem for Integrals }} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& {f_{avg}} = \frac{1}{{1 - \left( { - 1} \right)}}\int_{ - 1}^1 {\frac{{{x^2}}}{{{{\left( {{x^3} + 3} \right)}^2}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& {f_{avg}} = \frac{1}{{2\left( 3 \right)}}\int_{ - 1}^1 {\frac{{3{x^2}}}{{{{\left( {{x^3} + 3} \right)}^2}}}} dx \cr
& {f_{avg}} = \frac{1}{{2\left( 3 \right)}}\int_{ - 1}^1 {{{\left( {{x^3} + 3} \right)}^{ - 2}}\left( {3{x^2}} \right)} dx \cr
& {\text{Integrate using the general power rule for integration}} \cr
& {f_{avg}} = \frac{1}{{2\left( 3 \right)}}\left[ {\frac{{{{\left( {{x^3} + 3} \right)}^{ - 1}}}}{{ - 1}}} \right]_{ - 1}^1 \cr
& {f_{avg}} = - \frac{1}{6}\left[ {\frac{1}{{{x^3} + 3}}} \right]_{ - 1}^1 \cr
& {\text{Evaluate and simplify}} \cr
& {f_{avg}} = - \frac{1}{6}\left[ {\frac{1}{{{{\left( 1 \right)}^3} + 3}} - \frac{1}{{{{\left( { - 1} \right)}^3} + 3}}} \right] \cr
& {f_{avg}} = - \frac{1}{6}\left[ {\frac{1}{4} - \frac{1}{2}} \right] \cr
& {f_{avg}} = \frac{1}{{24}} \cr} $$
