Answer
$$\frac{{468}}{7}$$
Work Step by Step
$$\eqalign{
& \int_1^9 {x\root 3 \of {x - 1} dx} \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = x - 1,{\text{ }}x = u + 1,{\text{ }}dx = du \cr
& {\text{The new limits of integration are:}} \cr
& x = 9 \to u = 8 \cr
& x = 1 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_1^9 {x\root 3 \of {x - 1} dx} = \int_0^8 {\left( {u + 1} \right)\root 3 \of u du} \cr
& = \int_0^8 {\left( {u + 1} \right){u^{1/3}}du} \cr
& = \int_0^8 {\left( {{u^{4/3}} + {u^{1/3}}} \right)du} \cr
& = \left[ {\frac{3}{7}{u^{7/3}} + \frac{3}{4}{u^{4/3}}} \right]_0^8 \cr
& = \left[ {\frac{3}{7}{{\left( 8 \right)}^{7/3}} + \frac{3}{4}{{\left( 8 \right)}^{4/3}}} \right] - \left[ {\frac{3}{7}{{\left( 0 \right)}^{7/3}} + \frac{3}{4}{{\left( 0 \right)}^{4/3}}} \right] \cr
& {\text{Simplifying}} \cr
& = \frac{{468}}{7} - 0 \cr
& = \frac{{468}}{7} \cr} $$
