Answer
$$3\root 3 \of 2 + 6$$
Work Step by Step
$$\eqalign{
& \int_1^{11} {\frac{{dx}}{{{{\left( {x - 3} \right)}^{2/3}}}}} \cr
& {\text{The integrand is not continuous at }}x = 3,{\text{ then}} \cr
& \int_1^{11} {\frac{{dx}}{{{{\left( {x - 3} \right)}^{2/3}}}}} = \mathop {\lim }\limits_{a \to {3^ - }} \int_1^a {\frac{{dx}}{{{{\left( {x - 3} \right)}^{2/3}}}}} + \mathop {\lim }\limits_{b \to {3^ + }} \int_b^{11} {\frac{{dx}}{{{{\left( {x - 3} \right)}^{2/3}}}}} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{a \to {3^ - }} \left[ {3{{\left( {x - 3} \right)}^{1/3}}} \right]_1^a + \mathop {\lim }\limits_{b \to {3^ + }} \left[ {3{{\left( {x - 3} \right)}^{1/3}}} \right]_b^{11} \cr
& = 3\mathop {\lim }\limits_{a \to {3^ - }} \left[ {\root 3 \of {x - 3} } \right]_1^a + 3\mathop {\lim }\limits_{b \to {3^ + }} \left[ {\root 3 \of {x - 3} } \right]_b^{11} \cr
& = 3\mathop {\lim }\limits_{a \to {3^ - }} \left[ {\root 3 \of {a - 3} - \root 3 \of {1 - 3} } \right] + 3\mathop {\lim }\limits_{b \to {3^ + }} \left[ {\root 3 \of {11 - 3} - \root 3 \of {b - 3} } \right] \cr
& = 3\mathop {\lim }\limits_{a \to {3^ - }} \left[ {\root 3 \of {a - 3} + \root 3 \of 2 } \right] + 3\mathop {\lim }\limits_{b \to {3^ + }} \left[ {2 - \root 3 \of {b - 3} } \right] \cr
& {\text{Evaluating the limit}} \cr
& = 3\left[ {\root 3 \of {3 - 3} + \root 3 \of 2 } \right] + 3\left[ {2 - \root 3 \of {3 - 3} } \right] \cr
& = 3\root 3 \of 2 + 6 \cr} $$
