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?
 A: We could go for crude. Split into the integral from $1$ to $2$, and the integral from $2$ to $\infty$. 
On the interval $2$ to $\infty$ we have $x^3-1\gt x^3/4$, and therefore the easily computed $\int_2^\infty \frac{2}{x^{3/2}}\,dx$ provides an upper bound for that region.
On the interval $1$ to $2$, we have $x^3-1=(x-1)(x^2+x+1)\ge 3(x-1)$. And the integral $\int_1^2 \frac{1}{\sqrt{3(x-1)}}\,dx$ can be computed explicitly.
A: $$ \frac{1}{\sqrt{x^3-1}} = \frac{1}{\sqrt{x-1}\sqrt{x^2+x+1}} < \frac{1}{x\sqrt{x-1}}$$
$$ \int_1^\infty \frac{\mathrm{d}x}{x\sqrt{x-1}} = \pi < 4$$
This integral can be done by letting $u=\sqrt{x-1}$ which yields $\mathrm{d}x=2u\mathrm{d}u$,
$$ \int_0^\infty \frac{2 \mathrm{d}u}{u^2+1} = \pi$$
A: $$\begin{eqnarray*}\color{red}{I}=\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-1}}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+3x}}&=&\int_{0}^{+\infty}\frac{2\, dz}{\sqrt{z^4+3z^2+3}}\\&\color{red}{\leq}&\int_{0}^{+\infty}\frac{2\,dz}{\sqrt{z^4+3z^2+\frac{9}{4}}}=\color{red}{\pi\sqrt{\frac{2}{3}}.}\end{eqnarray*}$$
A tighter bound follows from Cauchy-Schwarz:

$$\begin{eqnarray*} \color{red}{I} &=&2\int_{0}^{+\infty}\frac{\sqrt{z^2+\sqrt{3}}}{\sqrt{z^4+3z^2+3}}\cdot\frac{dz}{\sqrt{z^2+\sqrt{3}}}\\&\color{red}{\leq}& 2\sqrt{\left(\int_{0}^{+\infty}\frac{z^2+\sqrt{3}}{z^4+3z^2+3}\,dz\right)\cdot\int_{0}^{+\infty}\frac{dz}{z^2+\sqrt{3}}}\\&=&2\pi\cdot\left(\frac{1}{6}-\frac{1}{4\sqrt{3}}\right)^{1/4}\leq\color{red}{\frac{2\pi}{42^{1/4}}}.\end{eqnarray*} $$

The manipulations in the first line show that $I$ is just twice a complete elliptic integral of the first kind, whose value can be computed through the arithmetic-geometric mean.
On the other hand, through the substitution $x=\frac{1}{t}$ and Euler's beta function we have:

$$ I \color{red}{=} \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}}\color{red}{\leq}\frac{3\cdot 6}{2\sqrt{3\pi}}=\sqrt{\frac{27}{\pi}}$$

since in a right neighbourhood of the origin we have $\Gamma(x)\leq\frac{1}{x}$.
As a by-product we get:

$$ \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}} = \frac{\pi}{\text{AGM}(\frac{1}{2} \sqrt{3+2 \sqrt{3}},3^{1/4})}$$

that allows us to compute $\Gamma\left(\frac{1}{6}\right)$ through an AGM-mean:

$$ \Gamma\left(\frac{1}{6}\right) = \color{red}{\frac{2^{\frac{14}{9}}\cdot 3^{\frac{1}{3}}\cdot \pi^{\frac{5}{6}} }{\text{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^{\frac{2}{3}}}}.$$

The last identity was missing in the Wikipedia page about particular values of the $\Gamma$ function, so I took the liberty to add it and add this answer as a reference.
A: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}}=\int_1^2 + \int_2^\infty=I_1+I_2.$
$I_2 = \int_2^\infty \frac{dx}{x^{3/2}\sqrt{1-1/x^3}}\color{red}{\leq}\lim_{A \to \infty}  \int_2^A \frac{dx}{x^{3/2}\sqrt{1-1/A^3}}=\lim_{A \to \infty}\sqrt{\frac{A^3}{A^3-1}}\int_2^A x^{-3/2}dx$
$I_1=\int_1^2\frac{dx}{(x-1)^{1/2}\sqrt{x^2+x+1}}\le \frac{1}{\sqrt{3}}\int_1^2(x-1)^{-1/2}d(x-1)$
