Proving an estimate for this integral How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$
 A: In the interval from $1$ to $2$, $\frac{(1+x)^{1/3}}{x^2}\le \frac{3^{1/3}}{x^2}$. 
Thus
$$\int_1^2 \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_1^2\frac{3^{1/3}}{x^2}\,dx= \frac{1}{2}(3^{1/3}).$$
In the interval from $2$ to $\infty$, $(1+x)^{1/3}\le \left(\frac{x}{2}+x\right)^{1/3}$. So 
$$\int_2^\infty \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_2^\infty(3/2)^{1/3}x^{-5/3}\,dx= \frac{3}{4}(3^{1/3}).$$
So our full integral is $\le \frac{5}{4}(3^{1/3})$. Since $(5/4)^3 \lt 2$, the result follows.  
A: $$
    \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x \stackrel{x=\frac{2-u}{u}}{=} 2^{4/3} \int_0^1 u^{-1/3} \left(2-u\right)^{-2} \mathrm{d} u = 2^{-2/3} \int_0^1 u^{-1/3} \left(1-\frac{1}{2} u\right)^{-2} \mathrm{d} u
$$
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.
A: The integral can be written as
$$\int_1^{\infty} {1 \over x^{5 \over 3}} \bigg(1 + {1 \over x}\bigg)^{1 \over 3}\,dx$$
By Taylor expanding, $(1 + {1 \over x})^{1 \over 3} = 1 + {1 \over 3x} +$ error term, where the error term is negative by the Lagrange form of the remainder. So the integral is less than
$$\int_1^{\infty} {1 \over x^{5 \over 3}} + {1 \over 3 x^{8 \over 3}}\,dx$$
$$= {3 \over 2} + {1 \over 3}\cdot{3 \over 5}$$
$$= 1.7$$
$$ < \sqrt[3]6$$
A: The given integral ($=:Q$) can be evaluated in elementary terms. Using the substitution $u:=(1+x)^{1/3}$ one obtains
$$Q=\int_{2^{1/3}}^\infty {3u^3\over (u^3-1)^2}\ du\ .$$
Mathematica produces the output

which numerically is $\doteq 1.66959$ (cf. Sasha's answer), whereas $6^{1/3}\doteq1.81712$.
A: Let $p=3$, $q=3/2$. Then 
\begin{align}
\int_1^{+\infty}\frac{(1+x)^{1/3}}{x^2}dx&=\int_1^{+\infty}\color{green}{\frac{(1+x)^{1/3}}{x}}\color{red}{\frac 1x}dx\\
&< \color{green}{\left(\int_1^{+\infty}\frac{1+x}{x^3}dx\right)^{1/3} }\color{red}{\left(\int_1^{+\infty}x^{—3/2}dx\right)^{2/3}}\\
&=(1+\int_1^{+\infty}x^{-3}dx)^{1/3}\left(\left[-2x^{-1/2}\right]_1^{+\infty}\right)^{2/3}\\
&=(1+\left[-\frac 12x^{-2}\right]_1^{+\infty})^{1/3}2^{2/3}\\
&=\sqrt[3]{3/2\cdot 4}\\
&=6^{1/3}.
\end{align}
Note that the inequality is strict as we are not in equality case in Hölder's inequality.
