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) ?

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.
