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

$\frac{\pi}{12}$

Using the property $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$, the given integral

$I=\int\limits_{\pi / 6}^{\pi / 3} \frac{d x}{1+\tan ^5 x}=\int\limits_{\pi / 6}^{\pi / 3} \frac{d x}{1+\tan ^5\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}=\int\limits_{\pi / 6}^{\pi / 3} \frac{d x}{1+\cot ^5 x}$

Here $2 I=\int\limits_{\pi / 6}^{\pi / 3} d x \Rightarrow I=\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$

Hence (2) is the correct answer.