Answer
$$\frac{3}{5}{\left( {\frac{3}{4}} \right)^{5/2}}$$
Work Step by Step
$$\eqalign{
& \int_{1/8}^1 {{x^{ - 1/3}}{{\left( {1 - {x^{2/3}}} \right)}^{3/2}}} dx \cr
& {\text{use the substitution method }} \cr
& u = 1 - {x^{2/3}},\,\,\,\,du = - \frac{2}{3}{x^{ - 1/3}}dx,\,\,\,\,\,dx = - \frac{3}{2}{x^{1/3}}du \cr
& {\text{Then}}{\text{,}} \cr
& \int {{x^{ - 1/3}}{{\left( {1 - {x^{2/3}}} \right)}^{3/2}}} dx = \int {{x^{ - 1/3}}{u^{3/2}}} \left( { - \frac{3}{2}{x^{1/3}}du} \right) \cr
& = - \frac{3}{2}\int {{u^{3/2}}} du \cr
& {\text{integrate}} \cr
& = - \frac{3}{2}\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) + C \cr
& = - \frac{3}{5}{u^{5/2}} + C \cr
& {\text{write in terms of }}r;{\text{ replace }}u = 1 - {x^{2/3}} \cr
& = - \frac{3}{5}{\left( {1 - {x^{2/3}}} \right)^{5/2}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_{1/8}^1 {{x^{ - 1/3}}{{\left( {1 - {x^{2/3}}} \right)}^{3/2}}} dx = - \frac{3}{5}\left( {{{\left( {1 - {x^{2/3}}} \right)}^{5/2}}} \right)_{1/8}^1 \cr
& {\text{Evaluating, we get:}} \cr
& = - \frac{3}{5}\left( {{{\left( {1 - {{\left( 1 \right)}^{2/3}}} \right)}^{5/2}} - {{\left( {1 - {{\left( {1/8} \right)}^{2/3}}} \right)}^{5/2}}} \right) \cr
& = - \frac{3}{5}\left( {{{\left( 0 \right)}^{5/2}} - {{\left( {3/4} \right)}^{5/2}}} \right) \cr
& = \frac{3}{5}{\left( {\frac{3}{4}} \right)^{5/2}} \cr} $$
