Answer
$$\frac{{1209}}{{28}}$$
Work Step by Step
$$\eqalign{
& \int_0^7 {x\root 3 \of {x + 1} dx} \cr
& {\text{Integrating }}\int {x\root 3 \of {x + 1} dx} \cr
& u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du \cr
& \int {x\root 3 \of {x + 1} dx} = \int {\left( {u - 1} \right)\root 3 \of u } du \cr
& = \int {\left( {{u^{4/3}} - {u^{1/3}}} \right)} du \cr
& = \frac{3}{7}{u^{7/3}} - \frac{3}{4}{u^{4/3}} + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = x + 1 \cr
& = \frac{3}{7}{\left( {x + 1} \right)^{7/3}} - \frac{3}{4}{\left( {x + 1} \right)^{4/3}} + C \cr
& {\text{Then,}} \cr
& \int_0^7 {x\root 3 \of {x + 1} dx} = \left[ {\frac{3}{7}{{\left( {x + 1} \right)}^{7/3}} - \frac{3}{4}{{\left( {x + 1} \right)}^{4/3}}} \right]_0^7 \cr
& = \left[ {\frac{3}{7}{{\left( {7 + 1} \right)}^{7/3}} - \frac{3}{4}{{\left( {7 + 1} \right)}^{4/3}}} \right] - \left[ {\frac{3}{7}{{\left( {0 + 1} \right)}^{7/3}} - \frac{3}{4}{{\left( {0 + 1} \right)}^{4/3}}} \right] \cr
& = \frac{{300}}{7} + \frac{9}{{28}} \cr
& = \frac{{1209}}{{28}} \cr} $$
