1. Let $I=\int_0^\pi \frac{(x+3)\sin x}{1+3\cos^2 x}\,dx.$

We split it as $I=\int_0^\pi \frac{x\sin x}{1+3\cos^2 x}\,dx+3\int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx.$

So write $I=I_1+3I_2,$ where $I_1=\int_0^\pi \frac{x\sin x}{1+3\cos^2 x}\,dx, \qquad I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx.$

2. Evaluate $I_1$ using the symmetry property If $f(x)=\frac{\sin x}{1+3\cos^2 x},$ then $f(\pi-x)=\frac{\sin(\pi-x)}{1+3\cos^2(\pi-x)}=\frac{\sin x}{1+3\cos^2 x}=f(x).$ So $f(x)$ is symmetric about $x=\frac\pi2$.

Now use the standard result $\int_0^\pi x f(x)\,dx=\frac\pi2\int_0^\pi f(x)\,dx$ whenever $f(\pi-x)=f(x)$.

Hence, $I_1=\frac\pi2\int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx=\frac\pi2 I_2.$

Therefore, $I=\left(\frac\pi2+3\right)I_2.$

3. Evaluate $I_2$ $I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx.$

Put $u=\cos x \quad \Rightarrow \quad du=-\sin x\,dx.$ When $x=0$, $u=1$; when $x=\pi$, $u=-1$.

Thus, $I_2=\int_1^{-1} \frac{-du}{1+3u^2}=\int_{-1}^1 \frac{du}{1+3u^2}.$

Now, $\int \frac{du}{1+3u^2}=\frac{1}{\sqrt3}\tan^{-1}(\sqrt3\,u).$

So, $I_2=\frac1{\sqrt3}\left[\tan^{-1}(\sqrt3\,u)\right]_{-1}^{1}$ $=\frac1{\sqrt3}\left(\tan^{-1}(\sqrt3)-\tan^{-1}(-\sqrt3)\right).$

Since $\tan^{-1}(\sqrt3)=\frac\pi3, \qquad \tan^{-1}(-\sqrt3)=-\frac\pi3,$ we get $I_2=\frac1{\sqrt3}\left(\frac\pi3+\frac\pi3\right)=\frac{2\pi}{3\sqrt3}.$

4. Substitute into $I$ $I=\left(\frac\pi2+3\right)\frac{2\pi}{3\sqrt3}.$

Simplify: $I=\frac{2\pi}{3\sqrt3}\cdot\frac{\pi+6}{2}$ $=\frac{\pi(\pi+6)}{3\sqrt3}.$

Hence, $\boxed{I=\frac{\pi}{3\sqrt3}(\pi+6)}.$

5. Compare with options This matches Option B: $\boxed{\frac{\pi}{3\sqrt3}(\pi+6)}.$