Proving an estimate for this integral How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$

$$
    \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x \stackrel{x=\frac{2-u}{u}}{=} 2^{4/3} \int_0^1 u^{-1/3} \left(2-u\right)^{-2} \mathrm{d} u = 2^{-2/3} \int_0^1 u^{-1/3} \left(1-\frac{1}{2} u\right)^{-2} \mathrm{d} u
$$
This is just the Euler type integral for Gauss's hypergeometric function:
$$
    \frac{\Gamma(a) \Gamma(c-a)} {\Gamma(c)} {}_2F_1\left(a,b;c;x\right) = \int_0^1 u^{a-1} (1-u)^{c-a-1} (1-x u)^{-b} \mathrm{d} u
$$
where $a=\frac{2}{3}$, $c=a+1 = \frac{5}{3}$, $b=2$ and $x=\frac{1}{2}$. Since $c=a+1$ the ratio of $\Gamma$-functions simplifies to $\frac{1}{a} = \frac{3}{2}$. Now
$$
    \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \frac{3}{2^{5/3}} {}_2F_1\left(\frac{2}{3}, 2; \frac{5}{3}; \frac{1}{2} \right)
$$
The said hypergeometric can be computed in (not very simple, but elementary) closed form:
$$
   \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \sqrt[3]{2}+\frac{1}{24} \log \left(3505753+2782518 \sqrt[3]{2}+2208486\
   2^{2/3}\right)-\frac{1}{2 \sqrt{3}}\arcsin\left(\frac{\sqrt{3}}{2} \sqrt{\left(7-3 \sqrt[3]{2}\right) \left(\sqrt[3]{2}-1\right)}\right) \approx 1.6695914 \approx 4.6540453^{1/3}
$$
So $6^{1/3}$ strikes me as a pretty tight bound.
