**Step 1: Analyze the Integral** Let $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(|x|+\frac{\pi}{6})}dx$. Since the interval of integration is $[\frac{\pi}{6}, \pi]$, $x$ is always positive, so $|x| = x$. Thus, $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$.

**Step 2: Apply the Property of Definite Integrals** Use the property $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$. $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4(\frac{7\pi}{6}-x)^{11}}{1-\sin(\frac{7\pi}{6}-x+\frac{\pi}{6})}dx = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4(\frac{7\pi}{6}-x)^{11}}{1-\sin(\frac{4\pi}{3}-x)}dx$.

**Step 3: Simplify the Expression** This substitution doesn't seem to simplify the integral directly. Let's try a different approach. Let $f(x) = \frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}$. Consider $I = \int_{\pi/6}^{\pi} \frac{\pi}{1-\sin(x+\pi/6)} dx + \int_{\pi/6}^{\pi} \frac{4x^{11}}{1-\sin(x+\pi/6)} dx$.

**Step 4: Use King Property** Let $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, we have $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4(\frac{7\pi}{6}-x)^{11}}{1-\sin(\frac{7\pi}{6}-x+\frac{\pi}{6})}dx = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4(\frac{7\pi}{6}-x)^{11}}{1-\sin(\frac{4\pi}{3}-x)}dx$. This doesn't seem to simplify things. Let's try another approach. $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi}{1-\sin(x+\frac{\pi}{6})}dx + \int_{\frac{\pi}{6}}^{\pi}\frac{4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$. Let $I_1 = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi}{1-\sin(x+\frac{\pi}{6})}dx$ and $I_2 = \int_{\frac{\pi}{6}}^{\pi}\frac{4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$.

**Step 5: Rationalize the denominator** $I_1 = \pi \int_{\frac{\pi}{6}}^{\pi} \frac{1+\sin(x+\frac{\pi}{6})}{1-\sin^2(x+\frac{\pi}{6})} dx = \pi \int_{\frac{\pi}{6}}^{\pi} \frac{1+\sin(x+\frac{\pi}{6})}{\cos^2(x+\frac{\pi}{6})} dx = \pi \int_{\frac{\pi}{6}}^{\pi} [\sec^2(x+\frac{\pi}{6}) + \sec(x+\frac{\pi}{6})\tan(x+\frac{\pi}{6})] dx$. $I_1 = \pi [\tan(x+\frac{\pi}{6}) + \sec(x+\frac{\pi}{6})]_{\frac{\pi}{6}}^{\pi} = \pi [(\tan(\frac{7\pi}{6}) + \sec(\frac{7\pi}{6})) - (\tan(\frac{\pi}{3}) + \sec(\frac{\pi}{3}))] = \pi [(\frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}) - (\sqrt{3} + 2)] = \pi [-\frac{1}{\sqrt{3}} - \sqrt{3} - 2] = \pi [-\frac{4}{\sqrt{3}} - 2]$. This is not leading to a simple answer.

**Step 6: Re-evaluate the approach** Let's consider $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$. Let $x = \frac{\pi}{6} + t$, then $dx = dt$. When $x = \frac{\pi}{6}$, $t = 0$. When $x = \pi$, $t = \frac{5\pi}{6}$. $I = \int_{0}^{\frac{5\pi}{6}}\frac{\pi+4(\frac{\pi}{6}+t)^{11}}{1-\sin(\frac{\pi}{6}+t+\frac{\pi}{6})}dt = \int_{0}^{\frac{5\pi}{6}}\frac{\pi+4(\frac{\pi}{6}+t)^{11}}{1-\sin(\frac{\pi}{3}+t)}dt$. This also doesn't seem to simplify.

**Step 7: Final Attempt** Let $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$. Let $f(x) = \frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}$. Let's try to find a function $g(x)$ such that $g'(x) = f(x)$. This seems difficult. Let's approximate the integral. The length of the interval is $\pi - \frac{\pi}{6} = \frac{5\pi}{6} \approx 2.618$. If we assume the average value of the function is around $4\pi$, then the integral would be approximately $4\pi$. Let's try to verify option (C) $6\pi$.

**Step 8: Consider the symmetry** Let $I = \int_{\frac{\pi}{6}}^{\pi}\frac{\pi+4x^{11}}{1-\sin(x+\frac{\pi}{6})}dx$. Let $u = x + \frac{\pi}{6}$. Then $x = u - \frac{\pi}{6}$ and $dx = du$. When $x = \frac{\pi}{6}$, $u = \frac{\pi}{3}$. When $x = \pi$, $u = \frac{7\pi}{6}$. $I = \int_{\frac{\pi}{3}}^{\frac{7\pi}{6}}\frac{\pi+4(u-\frac{\pi}{6})^{11}}{1-\sin(u)}du$. This also doesn't seem to simplify.

**Step 9: Numerical Approximation (Not Recommended for JEE)** Since we are unable to find an analytical solution, we can consider numerical methods. However, this is not the intended approach for a JEE question. Given the options, we can try to estimate the value. The function is positive in the interval. The integral is likely to be around $6\pi$.