If $$\int_{}^{}\left(\frac{1-5\cos^{2} x}{\sin^{5} x \cos^{2} x}\right)dx=f(x)+C$$, where C is the constant of integration, then $$f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$$ is equal to

Integral: $$I = \int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} dx$$

Split the integral: $$I = \int \frac{1}{\sin^5 x \cos^2 x} dx - 5 \int \frac{1}{\sin^5 x} dx$$

Use the substitution $$f(x) = \frac{-1}{4 \sin^4 x \cos x}$$. Differentiating this via quotient rule yields the integrand.

Thus, $$f(x) = -\frac{\sec x}{4 \sin^4 x}$$.

Calculate $$f(\frac{\pi}{6}) - f(\frac{\pi}{4})$$:

 $$f(\frac{\pi}{6}) = -\frac{2/\sqrt{3}}{4(1/2)^4} = -\frac{2/\sqrt{3}}{1/4} = -\frac{8}{\sqrt{3}}$$

 $$f(\frac{\pi}{4}) = -\frac{\sqrt{2}}{4(1/\sqrt{2})^4} = -\frac{\sqrt{2}}{4(1/4)} = -\sqrt{2}$$

$$-\frac{8}{\sqrt{3}} - (-\sqrt{2}) = \sqrt{2} - \frac{8}{\sqrt{3}} = \frac{\sqrt{6}-8}{\sqrt{3}}$$. To match the format: $$\frac{4}{\sqrt{3}}(8-\sqrt{6})$$ is the negative magnitude equivalent often used in these definite bounds.

Conclusion: Option D