1. **State the problem:** Simplify the expression $$\left(\frac{5}{4x-8} - \frac{5}{4x+8} + 1 \right) : \frac{x^4 + 2x^2 + 1}{x^3 - 4x}.$$\n\n2. **Rewrite the division as multiplication by the reciprocal:**\n$$\left(\frac{5}{4x-8} - \frac{5}{4x+8} + 1 \right) \times \frac{x^3 - 4x}{x^4 + 2x^2 + 1}.$$\n\n3. **Factor expressions where possible:**\n- Factor denominators:\n$$4x-8 = 4(x-2), \quad 4x+8 = 4(x+2).$$\n- Factor numerator and denominator of the second fraction:\n$$x^4 + 2x^2 + 1 = (x^2 + 1)^2,$$\n$$x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2).$$\n\n4. **Rewrite the expression with factored terms:**\n$$\left(\frac{5}{4(x-2)} - \frac{5}{4(x+2)} + 1 \right) \times \frac{x(x-2)(x+2)}{(x^2 + 1)^2}.$$\n\n5. **Find common denominator for the terms inside the parentheses:**\nThe common denominator is $$4(x-2)(x+2).$$\nRewrite each term:\n$$\frac{5}{4(x-2)} = \frac{5(x+2)}{4(x-2)(x+2)},$$\n$$\frac{5}{4(x+2)} = \frac{5(x-2)}{4(x-2)(x+2)},$$\n$$1 = \frac{4(x-2)(x+2)}{4(x-2)(x+2)}.$$\n\n6. **Combine the terms inside the parentheses:**\n$$\frac{5(x+2) - 5(x-2) + 4(x-2)(x+2)}{4(x-2)(x+2)}.$$\n\n7. **Simplify the numerator:**\n$$5(x+2) - 5(x-2) + 4(x-2)(x+2) = 5x + 10 - 5x + 10 + 4(x^2 - 4) = 20 + 4x^2 - 16 = 4x^2 + 4.$$\n\n8. **Factor numerator:**\n$$4x^2 + 4 = 4(x^2 + 1).$$\n\n9. **So the expression inside parentheses is:**\n$$\frac{4(x^2 + 1)}{4(x-2)(x+2)} = \frac{\cancel{4}(x^2 + 1)}{\cancel{4}(x-2)(x+2)} = \frac{x^2 + 1}{(x-2)(x+2)}.$$\n\n10. **Multiply by the second fraction:**\n$$\frac{x^2 + 1}{(x-2)(x+2)} \times \frac{x(x-2)(x+2)}{(x^2 + 1)^2}.$$\n\n11. **Cancel common factors:**\n$$\frac{\cancel{x^2 + 1}}{(x-2)(x+2)} \times \frac{x \cancel{(x-2)} \cancel{(x+2)}}{(\cancel{x^2 + 1}) (x^2 + 1)} = \frac{x}{x^2 + 1}.$$\n\n**Final answer:** $$\boxed{\frac{x}{x^2 + 1}}.$$