Answer
$$\frac{{3\left( {81 - {2^{4/3}}} \right)}}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^5 {x\root 3 \of {{x^2} + 2} } dx \cr
& {\text{use substitution}}{\text{. Let }}u = {x^2} + 2,{\text{ so that }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\,xdx = \frac{1}{2}du \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 5,{\text{ then }}u = {\left( 5 \right)^2} + 2 = 27 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = {\left( 0 \right)^2} + 2 = 2 \cr
& {\text{Then}} \cr
& \int_0^5 {x\root 3 \of {{x^2} + 2} } dx = \int_2^{27} {\root 3 \of u } \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_2^{27} {\root 3 \of u } du \cr
& {\text{write }}\root 3 \of u {\text{ as }}{u^{1/3}} \cr
& = \frac{1}{2}\int_2^{27} {{u^{1/3}}} du \cr
& {\text{integrate by the power rule}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{4/3}}}}{{4/3}}} \right)_2^{27} \cr
& = \frac{3}{8}\left( {{u^{4/3}}} \right)_2^{27} \cr
& {\text{evaluating}} \cr
& = \frac{3}{8}\left( {{{\left( {27} \right)}^{4/3}} - {{\left( 2 \right)}^{4/3}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{3}{8}\left( {81 - {2^{4/3}}} \right) \cr
& = \frac{{3\left( {81 - {2^{4/3}}} \right)}}{8} \cr} $$
