Step 1: Set up the inner ratio.
We are given $f(x)=\sqrt{\dfrac{x^2+2x+8}{x^2+2x+4}}$. Notice both quadratics share the same first two terms, so let us write everything in terms of a single quantity.
Step 2: Complete the square once.
Since $x^2+2x=(x+1)^2-1$, the numerator becomes $(x+1)^2+7$ and the denominator becomes $(x+1)^2+3$.
Step 3: Replace by a non-negative letter.
Put $u=(x+1)^2$, so $u\ge 0$. Then the inside is $g(u)=\dfrac{u+7}{u+3}$.
Step 4: Study how $g(u)$ behaves.
Write $g(u)=\dfrac{(u+3)+4}{u+3}=1+\dfrac{4}{u+3}$. As $u$ runs from $0$ upward, $u+3$ runs from $3$ upward, so $\dfrac{4}{u+3}$ decreases from $\dfrac{4}{3}$ down toward $0$ but never reaches $0$.
Step 5: Get the range of the inside.
Hence $g(u)$ takes its largest value $1+\dfrac{4}{3}=\dfrac{7}{3}$ at $u=0$, and approaches $1$ from above as $u\to\infty$. So $1<g(u)\le \dfrac{7}{3}$.
Step 6: Take the square root.
Since the square root is increasing, $1<\sqrt{g(u)}\le \sqrt{\dfrac{7}{3}}$. Therefore the range of $f$ is the interval that is open at $1$ and closed at $\sqrt{7/3}$.
\[ \boxed{\left(1,\ \sqrt{\dfrac{7}{3}}\right]} \]