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$\dfrac{x^2 + 1}{x} + \dfrac{x}{x^2 + 1} = \dfrac{29}{10}\\ \text{Let }u = \dfrac{x}{x^2 + 1}.\\ u + \dfrac{1}{u} = \dfrac{29}{10}\\ 10u^2 + 10 - 29u = 0\\ (2u - 5)(5u - 2) = 0\\ u = \dfrac{5}{2} \text{ or } u = \dfrac{2}{5}\\\\ \dfrac{x}{x^2 + 1} \stackrel{(1)}{=} \dfrac{5}{2}\text{ or } \dfrac{x}{x^2 + 1} \stackrel{(2)}{=} \dfrac{2}{5} \\ \text{Consider equation }(1),\\ 5x^2 + 5 = 2x\\ 5x^2 - 2x + 5 = 0\\ x = \dfrac{2 \pm \sqrt{2^2 - 4(5)^2}}{2(5)}\\ x = \dfrac{1\pm\sqrt{-24}}{5}\\ \text{Consider equation }(2),\\ 2x^2 + 2 = 5x\\ 2x^2 - 5x + 2 = 0\\ (2x - 1)(x - 2) = 0\\ \text{All roots are purely real.}\\ \text{The required integer is }-24.\\$