Answer
$${\text{The integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{dx}}{{\root 3 \of {x - 1} }}} \cr
& {\text{The integrand is not defined for }}x = 1,{\text{ then}} \cr
& \int_1^\infty {\frac{{dx}}{{\root 3 \of {x - 1} }}} = \mathop {\lim }\limits_{a \to {1^ + }} \int_a^c {\frac{{dx}}{{\root 3 \of {x - 1} }}} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {\frac{{{dx}}}{{{\sqrt[3]{x-1}}}} } \cr
& {\text{Where }}c\,{\text{is a number in the interval }}\left( {1,\infty } \right) \cr
& {\text{Integrating}} \cr
& = \frac{3}{2}\mathop {\lim }\limits_{a \to {1^ + }} \left[ {{{\left( {x - 1} \right)}^{2/3}}} \right]_a^c + \frac{3}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {x - 1} \right)}^{2/3}}} \right]_c^b \cr
& = \frac{3}{2}\mathop {\lim }\limits_{a \to {1^ + }} \left[ {{{\left( {c - 1} \right)}^{2/3}} - {{\left( {a - 1} \right)}^{2/3}}} \right] + \frac{3}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {b - 1} \right)}^{2/3}} - {{\left( {c - 1} \right)}^{2/3}}} \right] \cr
& = \frac{3}{2}\mathop {\lim }\limits_{a \to {1^ + }} \left[ {{{\left( {c - 1} \right)}^{2/3}} - {{\left( {a - 1} \right)}^{2/3}}} \right] + \frac{3}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {b - 1} \right)}^{2/3}} - 1} \right] \cr
& {\text{Evaluating the limits}} \cr
& = \frac{3}{2}\left[ {{{\left( {c - 1} \right)}^{2/3}} - {{\left( {1 - 1} \right)}^{2/3}}} \right] + \frac{3}{2}\left[ \infty \right] \cr
& = \frac{3}{2}\left[ {{{\left( {c - 1} \right)}^{2/3}} - {{\left( {1 - 1} \right)}^{2/3}}} \right] + \frac{3}{2}\left[ \infty \right] \cr
& = \frac{3}{2}{\left( {c - 1} \right)^{2/3}} + \infty \cr
& = \infty \cr
& {\text{The integral diverges}} \cr} $$
