Answer
$$\frac{2}{3}{\sin ^{ - 1}}{x^{3/2}} + C $$
Work Step by Step
$$\eqalign{
& \int {\sqrt {\frac{x}{{1 - {x^3}}}} } dx \cr
& \int {\frac{{\sqrt x }}{{\sqrt {1 - {x^3}} }}} dx \cr
& {\text{We use the hint }}\cr
& u = {x^{3/2}},\,\,\,\,\,x = {u^{2/3}},\,\,\,\,\,dx = \frac{2}{3}{u^{ - 1/3}}du \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{\sqrt x }}{{\sqrt {1 - {x^3}} }}} dx = \int {\frac{{\sqrt {{u^{2/3}}} }}{{\sqrt {1 - {{\left( {{u^{2/3}}} \right)}^3}} }}} \left( {\frac{2}{3}{u^{ - 1/3}}} \right)du \cr
& {\text{Simplifying, we get:}} \cr
& = \int {\frac{{{u^{1/3}}}}{{\sqrt {1 - {u^2}} }}} \left( {\frac{2}{3}{u^{ - 1/3}}} \right)du \cr
& = \frac{2}{3}\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{Integrating}}{\text{, use }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = {\sin ^{ - 1}}x + C \cr
& = \frac{2}{3}{\sin ^{ - 1}}u + C \cr
& {\text{Write in terms of }}x{\text{; replace }}u = {x^{3/2}} \cr
& = \frac{2}{3}{\sin ^{ - 1}}{x^{3/2}} + C \cr} $$
