The value of the integral ∫_ π/6 ^ π/3 ((4 - ^2 x)/( ^4 x)) dx is:

The given integral is $I = \displaystyle\int_{\pi/6}^{\pi/3} \left(\dfrac{4 - \csc^2 x}{\cos^4 x}\right) dx$.

We can rewrite the integrand in terms of $\tan x and \sec x$. Using the identity $\csc^2 x = 1 + \cot^2 x = 1 + \dfrac{1}{\tan^2 x}$, we get:

$I = \displaystyle\int_{\pi/6}^{\pi/3} \left(4 - 1 - \dfrac{1}{\tan^2 x}\right) \sec^4 x \, dx$

$I = \displaystyle\int_{\pi/6}^{\pi/3} \left(3 - \dfrac{1}{\tan^2 x}\right) (1 + \tan^2 x) \sec^2 x \, dx$

Substitute $t = \tan x$, which gives $dt = \sec^2 x \, dx$.

The limits of integration change as follows:
When $x = \dfrac{\pi}{6}$, $t = \dfrac{1}{\sqrt{3}}$.
When $x = \dfrac{\pi}{3}$, $t = \sqrt{3}$.

The integral becomes:

$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3 - \dfrac{1}{t^2}\right) (1 + t^2) dt$

Expanding the integrand, we get:

$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3 + 3t^2 - \dfrac{1}{t^2} - 1\right) dt$

$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3t^2 + 2 - t^{-2}\right) dt$

Integrating with respect to $t$:

$I = \left[ t^3 + 2t + \dfrac{1}{t} \right]_{1/\sqrt{3}}^{\sqrt{3}}$

Substitute the upper limit $t = \sqrt{3}$:

\left( \sqrt{3} \right)^3 + 2\sqrt{3} + \dfrac{1}{\sqrt{3}} = 3\sqrt{3} + 2\sqrt{3} + \dfrac{\sqrt{3}}{3} = \dfrac{16\sqrt{3}}{3} = \dfrac{48}{3\sqrt{3}}$<br><br>Substitute the lower limit$t = \dfrac{1}{\sqrt{3}}$:<br><br>\left( \dfrac{1}{\sqrt{3}} \right)^3 + 2\left( \dfrac{1}{\sqrt{3}} \right) + \sqrt{3} = \dfrac{1}{3\sqrt{3}} + \dfrac{2}{\sqrt{3}} + \sqrt{3} = \dfrac{1 + 6 + 9}{3\sqrt{3}} = \dfrac{16}{3\sqrt{3}}$

Subtracting the lower limit value from the upper limit value:

$I = \dfrac{48}{3\sqrt{3}} - \dfrac{16}{3\sqrt{3}} = \dfrac{32}{3\sqrt{3}}$

Answer: $\dfrac{32}{3\sqrt{3}}$