The value of $$\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}}\left(\frac{\pi+4x^{11}}{1-\sin(|x|+\frac{\pi}{6})}\right)dx$$ is equal to :

$$\int_{-\pi/6}^{\pi/6}\frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)}dx$$

Split: $$\int \frac{\pi}{1-\sin(|x|+\pi/6)}dx + \int \frac{4x^{11}}{1-\sin(|x|+\pi/6)}dx$$.

The second integral: $$x^{11}$$ is odd and $$1-\sin(|x|+\pi/6)$$ is even, so the integrand is odd → integral = 0.

First integral: $$\pi \int_{-\pi/6}^{\pi/6}\frac{dx}{1-\sin(|x|+\pi/6)}$$. Since the integrand is even:

$$= 2\pi\int_0^{\pi/6}\frac{dx}{1-\sin(x+\pi/6)}$$

Let $$u = x + \pi/6$$: $$= 2\pi\int_{\pi/6}^{\pi/3}\frac{du}{1-\sin u}$$

$$\frac{1}{1-\sin u} = \frac{1+\sin u}{\cos^2 u} = \sec^2 u + \sec u \tan u$$

$$\int(\sec^2 u + \sec u \tan u)du = \tan u + \sec u$$

$$= 2\pi[\tan u + \sec u]_{\pi/6}^{\pi/3} = 2\pi[(\sqrt{3}+2) - (1/\sqrt{3}+2/\sqrt{3})]$$

$$= 2\pi[(\sqrt{3}+2) - (3/\sqrt{3})] = 2\pi[(\sqrt{3}+2) - \sqrt{3}] = 2\pi \times 2 = 4\pi$$

The answer is Option 3: $$4\pi$$.