$$\begin {aligned} \int _{0}^{2}\frac {dx}{\sqrt {1+x^{3}}} &=\frac {2}3\int _{1}^{3}\frac {dt}{(t^2-1)^{\frac {2}3}}\quad \quad \left (t=\sqrt {1+x^{3}},x=(t^{2}-1)^{\frac {1}3}\right )\\ &=\frac {2}3\int _{\frac {1}2}^{1}\left (\frac {u}2\right )^{\frac {2}3}\left (\frac {2}{u}-2\right )^{-\frac {2}3}\frac {2du}{u^{2}}\quad \quad \left (u=\frac {2}{t+1},t=\frac {2}u-1\right )\\ &=\frac {2^{\frac {2}3}}3\int _{\frac {1}2}^{1}u^{-\frac {2}3}(1-u)^{-\frac {2}3}du\\ &=\frac {1}{3\sqrt[3] {2}}\int _{0}^{1}u^{{-\frac {2}3}}(1-u)^{-\frac {2}3}du\\ &=\frac {1}{3\sqrt[3] {2}}\cdot \frac {\Gamma \left (\frac {1}3\right )\Gamma \left (\frac {1}3\right )}{\Gamma \left (\frac {2}3\right )}\\ &=\frac {\Gamma \left (\frac {1}3\right )^3}{3\sqrt[3] {2}}\cdot \frac {1}{\Gamma \left (\frac {2}3\right )\Gamma \left (\frac {1}3\right )}\\ &=\frac {\Gamma \left (\frac {1}3\right )^3}{3\sqrt[3] {2}}\cdot \frac {\sin \frac {\pi }3}{\pi }\\ &=\textcolor {blue}{\frac {\Gamma \left (\frac {1}3\right )^3}{2^{\frac {4}3}\sqrt 3\pi }}. \end {aligned}$$