Answer
$$\frac{{{{\left( {t + 10} \right)}^{4/3}}}}{{14}}\left( {6t - 45} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {t\root 3 \of {t + 10} } dt \cr
& {\text{Integrate by substitution}} \cr
& u = t + 10,{\text{ }}t = u - 10,{\text{ }}dt = du \cr
& {\text{Substituting}} \cr
& \int {t\root 3 \of {t + 10} } dt = \int {\left( {u - 10} \right)\root 3 \of u } du \cr
& = \int {\left( {{u^{4/3}} - 10{u^{1/3}}} \right)du} \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{{{u^{7/3}}}}{{7/3}} - 10\left( {\frac{{{u^{4/3}}}}{{4/3}}} \right) + C \cr
& = \frac{3}{7}{u^{7/3}} - \frac{{30}}{4}{u^{4/3}} + C \cr
& = \frac{3}{7}{u^{7/3}} - \frac{{15}}{2}{u^{4/3}} + C \cr
& {\text{Factoring}} \cr
& = {u^{4/3}}\left( {\frac{3}{7}u - \frac{{15}}{2}} \right) + C \cr
& {\text{Write in terms of }}t,{\text{ }}u = t + 10 \cr
& = {\left( {t + 10} \right)^{4/3}}\left( {\frac{3}{7}\left( {t + 10} \right) - \frac{{15}}{2}} \right) + C \cr
& = {\left( {t + 10} \right)^{4/3}}\left( {\frac{3}{7}t + \frac{{30}}{7} - \frac{{15}}{2}} \right) + C \cr
& = {\left( {t + 10} \right)^{4/3}}\left( {\frac{3}{7}t - \frac{{45}}{{14}}} \right) + C \cr
& = \frac{{{{\left( {t + 10} \right)}^{4/3}}}}{{14}}\left( {6t - 45} \right) + C \cr} $$
