Showing $8.9<\int_3^5 \sqrt{4+x^2} \, \mathrm d x < 9$

Let $f(x)=\sqrt{4+x^2}$. Then $f$ is convex in $[3,5]$ and so the integral $I$ is less than area of the trapezoid above the graph: $$ I < \frac{f(3)+f(5)}{2}\cdot (5-3) = f(3)+f(5) = \sqrt{13}+\sqrt{29} \approx 8.99 < 9 $$

For $\,3<x<5\,$ we have $\,3.55+0.9(x-3)<\sqrt{4+x^2}<3.61+0.89(x-3)\,$ .

It's easily proofed by setting $\,x:=z+4\,$ with $\,-1<z<1\,$ and by the inequality conversion to

$-0.0321-0.01(1+z)-0.2079(1-z^2)<0<0.1875+0.01(1-z)+0.19z^2\,$.

Integration for $\,x\,$ from $\,3\,$ to $\,5\,$ proofs the assumption.