Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{6 x + \left(x^{3} + 1\right)}{x^{2} + 5 x}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{6 x + \left(x^{3} + 1\right)}{x^{2} + 5 x}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{3} + 6 x + 1}{x \left(x + 5\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{3} + 6 x + 1}{x \left(x + 5\right)}\right) = $$
$$\frac{1 + 3 \cdot 6 + 3^{3}}{3 \left(3 + 5\right)} = $$
= 23/12
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{6 x + \left(x^{3} + 1\right)}{x^{2} + 5 x}\right) = \frac{23}{12}$$