If $$x + \frac{1}{x} = 2\sqrt{5}$$, then what is the value of $$\frac{\left(x^4 + \frac{1}{x^2}\right)}{x^2 + 1}$$?

$$x+\frac{1}{x}=2\sqrt{5}$$...........(1)

$$\left(x+\frac{1}{x}\right)^3=40\sqrt{5}$$

$$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=40\sqrt{5}$$

$$x^3+\frac{1}{x^3}+3\left(2\sqrt{5}\right)=40\sqrt{5}$$  [From (1)]

$$x^3+\frac{1}{x^3}+6\sqrt{5}=40\sqrt{5}$$

$$x^3+\frac{1}{x^3}=34\sqrt{5}$$.........(2)

$$\frac{\left(x^4+\frac{1}{x^2}\right)}{x^2+1}=\frac{x\left(x^3+\frac{1}{x^3}\right)}{x\left(x+\frac{1}{x}\right)}$$

$$=\frac{x^3+\frac{1}{x^3}}{x+\frac{1}{x}}$$

$$=\frac{34\sqrt{5}}{2\sqrt{5}}$$

$$=17$$

Hence, the correct answer is Option B