Answer
$$A = \frac{1}{6}$$
Work Step by Step
$$\eqalign{
& y = {x^5} - 2\ln \left( {x + 5} \right),\,\,\,\,\,\,y = {x^3} - 2\ln \left( {x + 5} \right) \cr
& {\text{Find the points of intersection of the graphs equating the functions}} \cr
& {x^5} - 2\ln \left( {x + 5} \right) = {x^3} - 2\ln \left( {x + 5} \right) \cr
& {x^5} = {x^3} \cr
& {x^5} - {x^3} = 0 \cr
& {x^3}\left( {{x^2} - 1} \right) \cr
& {x^3}\left( {x + 1} \right)\left( {x - 1} \right) \cr
& x = 0,\,\,x = - 1,\,\,\,x = 1 \cr
& {\text{Then with the critical value we have the intervals }}\left( { - 1,0} \right){\text{ and }}\left( {0,1} \right) \cr
& {\text{For the interval }}\left( { - 1,0} \right){\text{ we have that }}{x^5} - 2\ln \left( {x + 5} \right) > {x^3} - 2\ln \left( {x + 5} \right) \cr
& {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^3} - 2\ln \left( {x + 5} \right) > {x^5} - 2\ln \left( {x + 5} \right) \cr
& \cr
& {\text{The area between the curves is given by}}: \cr
& A = \int_{ - 1}^0 {\left( {{x^5} - 2\ln \left( {x + 5} \right) - {x^3} + 2\ln \left( {x + 5} \right)} \right)} dx \cr
& + \int_0^1 {\left( {{x^3} - 2\ln \left( {x + 5} \right) - {x^5} + 2\ln \left( {x + 5} \right)} \right)} dx \cr
& \cr
& A = \int_{ - 1}^0 {\left( {{x^5} - {x^3}} \right)} dx + \int_0^1 {\left( {{x^3} - {x^5}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\frac{1}{6}{x^6} - \frac{1}{4}{x^4}} \right)_{ - 1}^0 + \left( {\frac{1}{4}{x^4} - \frac{1}{6}{x^6}} \right)_0^1 \cr
& A = \left( {\frac{1}{6}{{\left( 0 \right)}^6} - \frac{1}{4}{{\left( 0 \right)}^4}} \right) - \left( {\frac{1}{6}{{\left( { - 1} \right)}^6} - \frac{1}{4}{{\left( { - 1} \right)}^4}} \right) \cr
& + \left( {\frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{6}{{\left( 1 \right)}^6}} \right) - \left( {\frac{1}{4}{{\left( 0 \right)}^4} - \frac{1}{6}{{\left( 0 \right)}^6}} \right) \cr
& \cr
& A = \left( 0 \right) - \left( {\frac{1}{6} - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{6}} \right) - \left( 0 \right) \cr
& A = \frac{1}{6} \cr} $$
