Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$. 
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?

You're ever so slightly off. Notice the median of $(-5, -4, ..., 5, 6)$ is $\frac{-5+6}{2}=\frac 12$ which hints at trying $$f(\frac12+x)+f(\frac 12-x)\overbrace{=}^{y=x+\frac 12}f(y)+f(1-y)$$
We see that: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x}+2\cdot 3^1}=\frac{\alpha}{3^\frac 12 \cdot \alpha}=\frac{1}{\sqrt 3}$$
