1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{\ln\left(e^{3x} + x\right)}{x}.$$\n\n2. **Recall the properties and formula:** For large $x$, the term $e^{3x}$ grows much faster than $x$, so $e^{3x} + x \approx e^{3x}$. Also, recall that $\ln(a \cdot b) = \ln a + \ln b$ and $\ln(e^{k}) = k$.\n\n3. **Simplify the expression inside the logarithm:**\n$$\ln\left(e^{3x} + x\right) \approx \ln\left(e^{3x}\right) = 3x.$$\n\n4. **Rewrite the limit using this approximation:**\n$$\lim_{x \to \infty} \frac{\ln\left(e^{3x} + x\right)}{x} \approx \lim_{x \to \infty} \frac{3x}{x}.$$\n\n5. **Simplify the fraction:**\n$$\frac{3x}{x} = \cancel{\frac{3\cancel{x}}{\cancel{x}}} = 3.$$\n\n6. **Conclusion:** The limit is $$3.$$\n\nTherefore, $$\boxed{3}$$ is the value of the limit.