Question

$\frac{4x^{2} + x + 4}{x^{2} + 1} + \frac{x^{2} + 1}{x^{2} + x + 1} = \frac{31}{6}$

Let $\frac{4x^{2} + x + 4}{x^{2} + 1} = t \Rightarrow \frac{x^{2} + x + 1}{x^{2} + 1} = t - 3$

$$\begin{matrix} \Rightarrow & t + \left( \frac{1}{t - 3} \right) & = \frac{31}{6} \\ \Rightarrow & 6\left( t^{2} - 3t + 1 \right) & = 31t - 93 \\ \Rightarrow & 6t^{2} - 49t + 99 & = 0 \\ \Rightarrow & 6t^{2} - 27t - 22t + 99 & = 0 \\ \Rightarrow & (3t - 11)(2t - 9) & = 0 \\ \therefore & \frac{4x^{2} + x + 4}{x^{2} + 1} & = \frac{11}{3},\frac{9}{2} \\ \Rightarrow & x^{2} + 3x + 1 & = 0\text{~or~}x^{2} - 2x + 1 = 0 \\ & & x = \frac{- 3 \pm \sqrt{5}}{2},1 \end{matrix}$$