Answer
$$ - \frac{9}{2} + \frac{9}{{\root 3 \of 2 }}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^2 {\frac{3}{{{x^{1/3}}}}} dx \cr
& {\text{The integrand }}\frac{3}{{{x^{1/3}}}}{\text{ is not continuous to }}x = 0,{\text{ then}} \cr
& {\text{Write the integral as}} \cr
& \int_{ - 1}^2 {\frac{3}{{{x^{1/3}}}}} dx = \int_{ - 1}^0 {\frac{3}{{{x^{1/3}}}}} dx + \int_0^2 {\frac{3}{{{x^{1/3}}}}} dx \cr
& {\text{Improper Integral with an Infinite Limit of Integration}} \cr
& \int_{ - 1}^2 {\frac{3}{{{x^{1/3}}}}} dx = \mathop {\lim }\limits_{b \to {0^ - }} \int_{ - 1}^b {\frac{3}{{{x^{1/3}}}}} dx + \mathop {\lim }\limits_{a \to {0^ + }} \int_a^2 {\frac{3}{{{x^{1/3}}}}} dx \cr
& {\text{*Computing }}\mathop {\lim }\limits_{b \to {0^ - }} \int_{ - 1}^b {\frac{3}{{{x^{1/3}}}}} dx \cr
& \mathop {\lim }\limits_{b \to {0^ - }} \int_{ - 1}^b {\frac{3}{{{x^{1/3}}}}} dx = \mathop {\lim }\limits_{b \to {0^ - }} \left[ {\frac{{3{x^{2/3}}}}{{2/3}}} \right]_{ - 1}^b \cr
& {\text{ }} = \frac{9}{2}\mathop {\lim }\limits_{b \to {0^ - }} \left[ {{x^{2/3}}} \right]_{ - 1}^b \cr
& {\text{ }} = \frac{9}{2}\mathop {\lim }\limits_{b \to {0^ - }} \left[ {{b^{2/3}} - 1} \right] \cr
& {\text{ }} = \frac{9}{2}\left[ {{{\left( 0 \right)}^{2/3}} - 1} \right] \cr
& {\text{ }} = - \frac{9}{2} \cr
& {\text{*Computing }}\mathop {\lim }\limits_{a \to {0^ + }} \int_a^2 {\frac{3}{{{x^{1/3}}}}} dx \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \int_a^2 {\frac{3}{{{x^{1/3}}}}} dx = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{{3{x^{2/3}}}}{{2/3}}} \right]_a^2 \cr
& {\text{ }} = \frac{9}{2}\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{x^{2/3}}} \right]_a^2 \cr
& {\text{ }} = \frac{9}{2}\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{2^{2/3}} - {a^{2/3}}} \right] \cr
& {\text{ }} = \frac{9}{2}\left[ {{2^{2/3}} - 0} \right] \cr
& {\text{ }} = \frac{9}{{\root 3 \of 2 }} \cr
& {\text{Therefore,}} \cr
& \int_{ - 1}^2 {\frac{3}{{{x^{1/3}}}}} dx = - \frac{9}{2} + \frac{9}{{\root 3 \of 2 }} \cr} $$
