Answer
$$\frac{2}{3}\ln \left( {x\sqrt x + \sqrt {1 + {x^3}} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x }}{{\sqrt {1 + {x^3}} }}} dx \cr
& {\text{Let }}{u^{2/3}} = x,{\text{ }}\frac{2}{3}{u^{ - 1/3}}du = dx \cr
& {\text{Substitute}} \cr
& \int {\frac{{\sqrt x }}{{\sqrt {1 + {x^3}} }}} dx{\text{ }} = \int {\frac{{\sqrt {{u^{2/3}}} }}{{\sqrt {1 + {{\left( {{u^{2/3}}} \right)}^3}} }}\left( {\frac{2}{3}{u^{ - 1/3}}} \right)du} \cr
& = \int {\frac{{{u^{1/3}}}}{{\sqrt {1 + {u^2}} }}\left( {\frac{2}{3}{u^{ - 1/3}}} \right)du} \cr
& = \frac{2}{3}\int {\frac{1}{{\sqrt {1 + {u^2}} }}du} \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }} = } \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& = \frac{2}{3}\ln \left( {u + \sqrt {1 + {u^2}} } \right) + C \cr
& {\text{Write in terms of }}x,{\text{ }}{u^{2/3}} = x \to u = x\sqrt x \cr
& = \frac{2}{3}\ln \left( {x\sqrt x + \sqrt {1 + {{\left( {x\sqrt x } \right)}^2}} } \right) + C \cr
& = \frac{2}{3}\ln \left( {x\sqrt x + \sqrt {1 + {x^3}} } \right) + C \cr} $$
