I am trying to solve this integral, because it comes from the question: What happens if we extend intervals of some elliptic integral to all $\mathbb{R}$? And I found that some integrals converges.
My first work was substitute $u = x^3+1$ and I obtain $$\frac{1}{3}\int_{0}^{\infty}\frac{du}{\sqrt{u}(u-1)^{\frac{2}{3}}} = \frac{1}{3}\mathcal{M}\left(\frac{1}{2} \right) \left [ (u-1)^{-\frac{2}{3}} \right ]$$ After that I cant prove that this integral is equal to $\mathcal{B} \left ( \frac {1}{2},\frac {1}{3} \right )$.
$$\int_{-1}^{\infty}\frac{dx}{\sqrt{1+x^3}} = \int_{-1}^{0}\frac{dx}{\sqrt{1+x^3}} + \int_{0}^{\infty}\frac{dx}{\sqrt{1+x^3}}$$
$$\int_{0}^{\infty}\frac{dx}{\sqrt{1+x^3}} \xrightarrow[]{x=u^3} \frac{1}{3}\int_{0}^{\infty}\frac{u^{-\frac{2}{3}}}{\sqrt{1+u}}du = \frac{1}{3}\mathcal{B} \left(\frac{1}{3},\frac{1}{6} \right)$$
$$\int_{-1}^{0}\frac{dx}{\sqrt{1+x^3}} \xrightarrow[]{x=-u^3} \frac{1}{3}\int_{0}^{1}\frac{u^{-\frac{2}{3}}}{\sqrt{1-u}}du = \frac{1}{3}\mathcal{B} \left(\frac{1}{3},\frac{1}{2} \right)$$
$$\frac{1}{3}\mathcal{B} \left(\frac{1}{3},\frac{1}{2} \right) + \frac{1}{3}\mathcal{B} \left(\frac{1}{3},\frac{1}{6} \right) = \frac{\Gamma \left( \frac{1}{3} \right)}{3} \left(\frac{\Gamma \left( \frac{1}{6} \right) \Gamma \left( \frac{5}{6} \right) + \Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{1}{2} \right)}{\Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{5}{6} \right)} \right)$$ $$ \frac{\Gamma \left( \frac{1}{3} \right)}{3} \left(\frac{3 \Gamma \left( \frac{1}{2} \right)^2}{\Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{5}{6} \right)} \right) = \mathcal{B} \left(\frac{1}{3},\frac{1}{2} \right)$$