Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$ Background
I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.
Problem

Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$

Attempt
If we define $f(x) := \frac{1+\cos(x)}{2+\cos(x)}$, which most definitely is $\pi$ - periodic.
The integral is, using our notation above, on the form
$$I = \int_{-\infty}^\infty \frac{\sin(x)}{x}f(x) dx.$$
The integrand is even, so we might as well compute
$$ I = 2 \int_{0}^\infty \frac{\sin(x)}{x}f(x) dx.$$
We will now have to make use of a theorem.

Lobachevsky's integral formula states that if $f(x)$ is a continous $\pi$ - periodic function then we have that $$ \int_0^\infty \frac{\sin(x)}{x}f(x) dx= \int_0^{\pi/2} f(x) dx.$$

Substituing our $f(x)$ yields us
$$ \int_0^{\pi/2} \frac{1+\cos(x)}{2+\cos(x)} dx = \pi/2 - \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx $$
where
$$I_2 = \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx = \int_0^{\pi/2}\frac{\sec^2(x/2)}{3+\tan^2(x/2)}dx.$$
Letting $ u = \tan(x/2)/\sqrt{3}$, for which $du = \sec^2(x/2)/(2\sqrt{3})dx$, therefore gives us:
$$ I_2 = \int_0^{1/\sqrt{3}}\frac{2\sqrt{3}}{3u^2+3} = \frac{\pi}{3\sqrt{3}}.$$
Finally we can compute $I$ to
$$I = 2\left(\frac{\pi}{2} - \frac{\pi}{3\sqrt{3}}\right) = \frac{\pi(3\sqrt{3}-2)}{3\sqrt{3}}.$$
I've tried calculating this integral in Desmos where it gives me $0$ when I calculate the integrand on the interval $(-\infty, \infty)$, and something negative for $(0,\infty)$. This contradicts my answer.
I also tried typing it into Wolfram, without success. Can anyone confirm the validity of my result?

Utilize the series
$$\frac{p\sin x}{1+2p\cos x+p^2}=-\sum_{k=1}^\infty (-p)^{k}\sin kx
$$
with $p=2-\sqrt3$ to express
\begin{align}
\frac{\sin x}{2+\cos x}= -2\sum_{k=1}^\infty (\sqrt3-2)^k\sin kx
\end{align}
Then, integrate
\begin{align}
&\int_{-\infty}^\infty \frac{\sin x}x\frac{1+\cos x}{2+\cos x}dx\\
=& \ 2\int_{-\infty}^\infty \frac{\sin x}xdx- 2\int_{-\infty}^\infty \frac{1}x\frac{\sin x}{2+\cos x}dx\\
=&\ \pi +4\sum_{k=1}^\infty (\sqrt3-2)^k\int_0^\infty \frac{\sin kx}xdx\\
=&\ \pi + 2\pi \sum_{k=1}^\infty (\sqrt3-2)^k= \frac\pi{\sqrt3}
\end{align}
where the summation is geometric.
