1. We need to evaluate $I=\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\tan 2x}}.$

2. Use the substitution $t=\frac{\pi}{4}-x.$ Then $dx=-dt.$ Also, when $x=\frac{\pi}{24}$, $t=\frac{\pi}{4}-\frac{\pi}{24}=\frac{5\pi}{24},$ and when $x=\frac{5\pi}{24}$, $t=\frac\pi4-\frac{5\pi}{24}=\frac{\pi}{24}.$ So the interval maps onto itself in reverse.

Now, $\tan 2t=\tan\left(\frac\pi2-2x\right)=\cot 2x=\frac{1}{\tan 2x}.$ Hence $\sqrt[3]{\tan 2t}=\frac{1}{\sqrt[3]{\tan 2x}}.$

3. Let $f(x)=\frac{1}{1+\sqrt[3]{\tan 2x}}.$ Then under the transformation $x\mapsto \frac\pi4-x$, \begin{align*} f\left(\frac\pi4-x\right) &=\frac{1}{1+\sqrt[3]{\tan\left(\frac\pi2-2x\right)}}\ &=\frac{1}{1+\sqrt[3]{\cot 2x}}\ &=\frac{1}{1+\frac{1}{\sqrt[3]{\tan 2x}}}\ &=\frac{\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}}. \end{align*} Therefore, $f(x)+f\left(\frac\pi4-x\right)=\frac{1}{1+a}+\frac{a}{1+a}=1,$ where $a=\sqrt[3]{\tan 2x}$.

4. Since the limits satisfy $\frac{\pi}{24}+\frac{5\pi}{24}=\frac\pi4,$ we use the standard property $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx.$ Thus, $I=\int_{\pi/24}^{5\pi/24} f\left(\frac\pi4-x\right)dx.$ Adding this with the original integral, \begin{align*} 2I&=\int_{\pi/24}^{5\pi/24}\left[f(x)+f\left(\frac\pi4-x\right)\right]dx\ &=\int_{\pi/24}^{5\pi/24}1,dx\ &=\frac{5\pi}{24}-\frac{\pi}{24}\ &=\frac{4\pi}{24}=\frac\pi6. \end{align*} So, $I=\frac{\pi}{12}.$

5. Therefore the correct option is $\boxed{\text{C }\frac{\pi}{12}}.$