The value of the integral $\int\limits_{π/6}^{π/3}\frac{dx}{1+tan^5x}$ is

π/12

Using the property $\int\limits_{a}^{b}f(x) dx = \int\limits_{a}^{b}f(a + b – x) dx$, the given integral

$I=\int\limits_{π/6}^{π/3}\frac{dx}{1+tan^5x}=\int\limits_{π/6}^{π/3}\frac{dx}{1+tan^5(\frac{π}{3}+\frac{π}{6}-x)}=\int\limits_{π/6}^{π/3}\frac{dx}{1+cot^5x}=\int\limits_{π/6}^{π/3}\frac{tan^5dx}{1+tan^5x}$

Hence $2 I =\int\limits_{π/6}^{π/3}dx⇒I=\frac{1}{2}(\frac{π}{3}-\frac{π}{6})=\frac{π}{12}$.

Hence (B) is the correct answer.