Answer
$$\frac{9}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^9 {\frac{{dx}}{{{{\left( {x - 1} \right)}^{1/3}}}}} \cr
& {\text{The integrand is not continuous at }}x = 1,{\text{ then}} \cr
& \int_0^9 {\frac{{dx}}{{{{\left( {x - 1} \right)}^{1/3}}}}} = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {\frac{{dx}}{{{{\left( {x - 1} \right)}^{1/3}}}}} + \mathop {\lim }\limits_{b \to {1^ + }} \int_b^9 {\frac{{dx}}{{{{\left( {x - 1} \right)}^{1/3}}}}} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{a \to {1^ - }} \left[ {\frac{3}{2}{{\left( {x - 1} \right)}^{2/3}}} \right]_0^a + \mathop {\lim }\limits_{b \to {1^ + }} \left[ {\frac{3}{2}{{\left( {x - 1} \right)}^{2/3}}} \right]_b^9 \cr
& = \frac{3}{2}\mathop {\lim }\limits_{a \to {1^ - }} \left[ {{{\left( {a - 1} \right)}^{2/3}} - {{\left( {0 - 1} \right)}^{2/3}}} \right] + \frac{3}{2}\mathop {\lim }\limits_{b \to {1^ + }} \left[ {{{\left( {9 - 1} \right)}^{2/3}} - {{\left( {b - 1} \right)}^{2/3}}} \right] \cr
& = \frac{3}{2}\mathop {\lim }\limits_{a \to {1^ - }} \left[ {{{\left( {a - 1} \right)}^{2/3}} - 1} \right] + \frac{3}{2}\mathop {\lim }\limits_{b \to {1^ + }} \left[ {4 - {{\left( {b - 1} \right)}^{2/3}}} \right] \cr
& {\text{Evaluating the limit}} \cr
& = \frac{3}{2}\left[ {{{\left( {1 - 1} \right)}^{2/3}} - 1} \right] + \frac{3}{2}\left[ {4 - {{\left( {1 - 1} \right)}^{2/3}}} \right] \cr
& = \frac{3}{2}\left( { - 1} \right) + \frac{3}{2}\left( 4 \right) \cr
& = \frac{9}{2} \cr} $$
