Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{1}{{\root 3 \of {x - 1} }}} dx \cr
& \frac{1}{{\root 3 \of {x - 1} }}{\text{ has an infinite discontinuity at }}x = 1,{\text{ so we can write}} \cr
& \int_0^2 {\frac{1}{{\root 3 \of {x - 1} }}} dx = \mathop {\lim }\limits_{b \to {1^ - }} \int_0^b {\frac{1}{{\root 3 \of {x - 1} }}} dx + \mathop {\lim }\limits_{a \to {1^ + }} \int_a^2 {\frac{1}{{\root 3 \of {x - 1} }}} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to {1^ - }} \left[ {\frac{3}{2}{{\left( {x - 1} \right)}^{2/3}}} \right]_0^b + \mathop {\lim }\limits_{a \to {1^ + }} \left[ {\frac{3}{2}{{\left( {x - 1} \right)}^{2/3}}} \right]_a^2 \cr
& = \mathop {\lim }\limits_{b \to {1^ - }} \left[ {\frac{3}{2}{{\left( {b - 1} \right)}^{2/3}} - \frac{3}{2}{{\left( {0 - 1} \right)}^{2/3}}} \right] + \mathop {\lim }\limits_{a \to {1^ + }} \left[ {\frac{3}{2}{{\left( {2 - 1} \right)}^{2/3}} - \frac{3}{2}{{\left( {a - 1} \right)}^{2/3}}} \right]_a^2 \cr
& = \mathop {\lim }\limits_{b \to {1^ - }} \left[ {\frac{3}{2}{{\left( {b - 1} \right)}^{2/3}} - \frac{3}{2}} \right] + \mathop {\lim }\limits_{a \to {1^ + }} \left[ {\frac{3}{2} - \frac{3}{2}{{\left( {a - 1} \right)}^{2/3}}} \right]_a^2 \cr
& {\text{Evaluate limits}} \cr
& = \left[ {\frac{3}{2}{{\left( {1 - 1} \right)}^{2/3}} - \frac{3}{2}} \right] + \left[ {\frac{3}{2} - \frac{3}{2}{{\left( {1 - 1} \right)}^{2/3}}} \right] \cr
& = - \frac{3}{2} + \frac{3}{2} \cr
& = 0 \cr
& {\text{The following graph confirms the result}} \cr} $$
