If $$x + \frac{1}{x} = 5{\text{,}}$$   then the value of $$\frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} = ?$$

$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \left( {{\text{By squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = 23 \cr & {\text{Now,}} \cr & \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & {\text{Divided by }}{x^2}, \cr & \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr & \Rightarrow \frac{{43}}{{23}} \cr} $$