How to prove $\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$? How to prove the following? $$\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$$
Ok I know we can directly integrate the function.The answer is given by Wolfram Alpha 
But is there any shorter/elegant method to obtain the left and right bounds directly by squeeze theorem (i.e. by integrating two function in between which the given function lies) ?
 A: If $0\le x\le 1$ then $\dfrac 1 3 \le \dfrac 1 {2+1^2} \le \dfrac 1 {2+x^2} \le \dfrac 1 {2+0^2} = \dfrac 1 2$.
If $1\le x \le 2$ then $\dfrac 1 6 \le\dfrac 1 {2+x^2} \le \dfrac 1 3 $.
Therefore
$$
\frac 1 3 + \frac 1 6 \le \int_0^1 \frac{dx}{2+x^2} + \int_1^2 \frac{dx}{2+x^2} \le \frac 1 2 + \frac 1 3.
$$
As for Wolfram Alpha, notice that the integral is easily evaluated in closed form:
\begin{align}
& \int_0^2 \frac{dx}{2+x^2} = \frac 1 2 \int_0^2 \frac{dx}{1 + \left(\frac x {\sqrt 2}\right)^2} = \frac{\sqrt 2} 2 \int_0^2 \frac{dx/\sqrt 2}{1+\left( \frac x {\sqrt2} \right)^2} = \frac {\sqrt 2} 2 \int_0^{\sqrt 2} \frac{du}{1+u^2} \\[10pt]
= {} & \frac{\sqrt 2} 2 \left( \arctan \sqrt 2 - \arctan 0 \right) = \frac{\sqrt 2} 2 \arctan \sqrt 2.
\end{align}
A: On $[0,1]$ you get
$$
\frac13\le\frac1{2+x^2}\le\frac12
$$
and on $[1,2]$
$$
\frac16\le\frac1{2+x^2}\le\frac13
$$
The combination of the estimates for both intervals gives exactly the requested bound.
A: If we don't want use the closed form of the integral, we can see that $$\int_{0}^{2}\frac{dx}{2+x^{2}}\stackrel{x=2u}{=}\int_{0}^{1}\frac{du}{1+2u^{2}}
 $$ and $$\int_{0}^{2}\frac{dx}{2+x^{2}}=\int_{0}^{1}\frac{du}{1+2u^{2}}\geq\int_{0}^{1}\frac{du}{1+2u}=\frac{1}{2}\log\left(3\right)>\frac{1}{2},$$
 $$\int_{0}^{1}\frac{du}{1+2u^{2}}\leq\int_{0}^{1}\frac{du}{1+u^{2}}=\frac{\pi}{4}<\frac{5}{6}.$$
