Answer
$$\frac{5}{4}\left( {\root 5 \of {{3^4}} - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx \cr
& {\text{The integrand }}\frac{3}{{{x^{1/3}}}}{\text{ is not continuous to }}x = - 1,{\text{ then}} \cr
& {\text{Write the integral as}} \cr
& \int_{ - 2}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = \int_{ - 2}^{ - 1} {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx + \int_{ - 1}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx \cr
& {\text{Improper Integral with an Infinite Limit of Integration}} \cr
& \int_{ - 2}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = \mathop {\lim }\limits_{a \to - {1^ - }} \int_{ - 2}^a {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx + \mathop {\lim }\limits_{b \to - {1^ + }} \int_b^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx \cr
& {\text{*Computing }}\mathop {\lim }\limits_{a \to - {1^ - }} \int_{ - 2}^a {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx \cr
& \mathop {\lim }\limits_{a \to - {1^ - }} \int_{ - 2}^a {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = \mathop {\lim }\limits_{a \to - {1^ - }} \left[ {\frac{5}{4}{{\left( {x + 1} \right)}^{4/5}}} \right]_{ - 2}^a \cr
& = \mathop {\lim }\limits_{a \to - {1^ - }} \left[ {\frac{5}{4}{{\left( {a + 1} \right)}^{4/5}} - \frac{5}{4}{{\left( { - 2 + 1} \right)}^{4/5}}} \right] \cr
& = \mathop {\lim }\limits_{a \to - {1^ - }} \left[ {\frac{5}{4}{{\left( {a + 1} \right)}^{4/5}} - \frac{5}{4}} \right] \cr
& = \frac{5}{4}{\left( { - 1 + 1} \right)^{4/5}} - \frac{5}{4} \cr
& = - \frac{5}{4} \cr
& {\text{*Computing }}\mathop {\lim }\limits_{b \to - {1^ + }} \int_b^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx \cr
& \mathop {\lim }\limits_{b \to - {1^ + }} \int_b^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = \mathop {\lim }\limits_{b \to - {1^ + }} \left[ {\frac{5}{4}{{\left( {x + 1} \right)}^{4/5}}} \right]_b^2 \cr
& = \mathop {\lim }\limits_{b \to - {1^ + }} \left[ {\frac{5}{4}{{\left( {2 + 1} \right)}^{4/5}} - \frac{5}{4}{{\left( {b + 1} \right)}^{4/5}}} \right] \cr
& = \mathop {\lim }\limits_{b \to - {1^ + }} \left[ {\frac{5}{4}{{\left( 3 \right)}^{4/5}} - \frac{5}{4}{{\left( {b + 1} \right)}^{4/5}}} \right] \cr
& = \frac{5}{4}{\left( 3 \right)^{4/5}} - \frac{5}{4}{\left( { - 1 + 1} \right)^{4/5}} \cr
& = \frac{5}{4}{\left( 3 \right)^{4/5}} \cr
& {\text{Therefore,}} \cr
& \int_{ - 2}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = - \frac{5}{4} + \frac{5}{4}{\left( 3 \right)^{4/5}} \cr
& \int_{ - 2}^2 {\frac{1}{{{{\left( {x + 1} \right)}^{1/5}}}}} dx = \frac{5}{4}\left( {\root 5 \of {{3^4}} - 1} \right) \cr} $$
