Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(2 x + 1\right)}{x^{4} - 1}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(2 x + 1\right)}{x^{4} - 1}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 2 x + 1}{\left(x - 1\right) \left(x + 1\right) \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 2 x + 1}{x^{4} - 1}\right) = $$
$$\frac{1 + 2 \cdot 2 + 3 \cdot 2^{2}}{-1 + 2^{4}} = $$
= 17/15
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(2 x + 1\right)}{x^{4} - 1}\right) = \frac{17}{15}$$