Integral inequality: $\frac{1}{30}<\int_{2}^{\infty}\frac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\,ds<\frac{\sqrt{2}}{20}$ Prove that
$$\frac{1}{30}<\int_{2}^{\infty}\frac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\,ds<\frac{\sqrt{2}}{20}$$
I think this inequality can $s^5+s^2+1>( ) $
and $\sqrt{s^3-s^2+3}<()$?
 A: For right inequality, write
$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s < \int_{2}^{\infty}\dfrac{\sqrt{s^3}}{s^5}\mathrm{d}s = \int_{2}^{\infty}s^{-\frac{7}{2}} \mathrm{d}s=\frac{\sqrt{2}}{20}$$
For left one, you may try
$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s > \int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s$$
Then using the fact that $s^5+s^3+s^2+1 = (s+1)(s^2+1)(s^2-s+1)$, you have
$$\frac{1}{s^5+s^3+s^2+1}=-\frac{2s-1}{3(s^2-s+1)}+\frac{s+1}{2(s^2+1)}+\frac{1}{6(s+1)}$$
Then you can compute the integral, 
$$\int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s = \sqrt{7} \left(-\frac{\log 5}{4}+\frac{\log 3}{6}+{1\over2}\arctan {1\over2}\right)$$
But it's a bit below $\frac{1}{30}$ ($\approx 0.997 \cdot \frac{1}{30}$) :-(
A: Denote your integral by $Q$. Then
$$Q=\int_2^\infty {1\over s^{7/2}}\ {\sqrt{1-{1\over s}+{3\over s^3}}\over 1+{1\over s^3}+{1\over s^5}}\ ds\ .$$
First off, one has
$$\int_2^\infty {1\over s^{7/2}}\ ds={\sqrt{2}\over 20}\ .$$
Furthermore one easily checks that in the interval $2\leq s<\infty$ the following estimates are valid:
$$\sqrt{1\over2}\leq\sqrt{1-{1\over s}+{3\over s^3}}=\sqrt{1-{1\over s}\bigl(1-{3\over s^2}\bigr)}\leq 1\ ,$$
$$1\leq 1+{1\over s^3}+{1\over s^5}\leq 1+{1\over 8}+{1\over32}={37\over32}\ .$$
It follows that
$${\sqrt{1/ 2}\over 37/32}\cdot {\sqrt{2}\over 20}\leq Q\leq {1\over 1}\cdot {\sqrt{2}\over 20}\ .$$
Since ${32\over 740}>{1\over30}$ the stated inequalities are indeed true.
