Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$ I'm going through Nahin's book Inside Interesting Integrals, and I'm stuck at an early problem, Challenge Problem 1.2: to show that
$$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}$$
exists because there is a finite upper-bound on its value. In particular, show that the integral is less than 4.
I've tried various substitutions and also comparisons with similar integrals, but the problem is that any other integral that I can easily compare the above integral to is just as hard to integrate, which doesn't help solve the problem.
I also tried just looking at the graph and hoping for insight, but that didn't work either.
So how doesone one place an upper bound on the integral?

$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}=\int_1^2 + \int_2^\infty=I_1+I_2.$
$I_2 = \int_2^\infty \frac{dx}{x^{3/2}\sqrt{1-1/x^3}}\color{red}{\leq}\lim_{A \to \infty}  \int_2^A \frac{dx}{x^{3/2}\sqrt{1-1/A^3}}=\lim_{A \to \infty}\sqrt{\frac{A^3}{A^3-1}}\int_2^A x^{-3/2}dx$
$I_1=\int_1^2\frac{dx}{(x-1)^{1/2}\sqrt{x^2+x+1}}\le \frac{1}{\sqrt{3}}\int_1^2(x-1)^{-1/2}d(x-1)$
