Answer
$$\frac{{4752}}{{35}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^6 {{x^2}\root 3 \of {x + 2} dx} \cr
& {\text{Integrating }}\int {x\root 3 \of {x + 1} dx} \cr
& u = x + 2,{\text{ }}x = u - 2,{\text{ }}dx = du \cr
& \int {{x^2}\root 3 \of {x + 1} dx} = \int {{{\left( {u - 2} \right)}^2}\root 3 \of u } du \cr
& = \int {\left( {{u^2} - 4u + 4} \right)} {u^{1/3}}du \cr
& = \int {\left( {{u^{7/3}} - 4{u^{4/3}} + 4{u^{1/3}}} \right)} du \cr
& = \frac{3}{{10}}{u^{10/3}} - 4\left( {\frac{3}{7}{u^{7/3}}} \right) + 4\left( {\frac{3}{4}{u^{4/3}}} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = x + 2 \cr
& = \frac{3}{{10}}{\left( {x + 2} \right)^{10/3}} - \frac{{12}}{7}{\left( {x + 2} \right)^{7/3}} + 3{\left( {x + 2} \right)^{4/3}} + C \cr
& {\text{Then,}} \cr
& \int_{ - 2}^6 {{x^2}\root 3 \of {x + 2} dx} = \left[ {\frac{3}{{10}}{{\left( {x + 2} \right)}^{10/3}} - \frac{{12}}{7}{{\left( {x + 2} \right)}^{7/3}} + 3{{\left( {x + 2} \right)}^{4/3}}} \right]_{ - 2}^5 \cr
& = \left[ {\frac{3}{{10}}{{\left( 8 \right)}^{10/3}} - \frac{{12}}{7}{{\left( 8 \right)}^{7/3}} + 3{{\left( 8 \right)}^{4/3}}} \right] - \left[ 0 \right] \cr
& = \frac{{4752}}{{35}} \cr} $$
