How to solve $\int_{1}^{\infty} \frac{1}{\sqrt{x^3 -1}}\,dx $ This integral converges because $$ \int_{1}^{\infty} \frac{1}{\sqrt{x^3 -1}}\,dx < \int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx  $$
$$\int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx = 2\int_{0}^{\infty} \frac{1}{u^2+1}\,dx =\pi$$
But, how to solve it?

Substitute $y=x^{-3}$ to recover a beta integral.
$$\int_1^\infty \frac{dx}{\sqrt{x^3-1}} = \int_1^\infty \frac{dx}{x^{3/2} \sqrt{1-x^{-3}}} \\
= \frac13 \int_0^1 \frac{y^{-4/3}\,dy}{y^{-1/2}\sqrt{1-y}}=\frac13 \int_0^1 y^{-5/6} (1-y)^{-1/2} \, dy$$
