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

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.  
