An algebra problem by Paola Ramírez

$\text{Consider } \quad 2x+1 = t \\ \dfrac{4x^2}{(1-\sqrt{1+2x})^2}<2x+9 \implies \dfrac{(t-1)^2}{(1-\sqrt{t})^2}<t+8 \\ D_{f} \rightarrow x\ge-\dfrac{1}{2} \text{ and } x\ne 0 \\ \dfrac{\cancel{(\sqrt{t}-1)^2}(\sqrt{t}+1)^2}{\cancel{(1-\sqrt{t})^2}} < t+8 \\ t+1+2\sqrt{t}<t+8 \\ \sqrt{t}<\dfrac{7}{2} \implies t<\dfrac{49}{4} \\ \boxed{x<\dfrac{45}{8} } \\ \text{Possible values of } x = 1,2,3,4,5 \\ \text{ Answer :} \quad \dfrac{5\cdot 6}{2} = \color{royalblue}{\boxed{15}}$