To determine the least integer $n$ such that $f(x) = 2x^2 + x^3 \log x$ is $O(x^n)$, we need to focus on the asymptotic growth of the function's terms as $x \to \infty$.

Step-by-Step Process:

1. Break the function into parts:

   • $f(x) = 2x^2 + x^3 \log x$
   • Here we have two terms: $2x^2$ and $x^3 \log x$.

2. Analyze the growth of each term:

   • $2x^2$: This term grows like $x^2$.
   • $x^3 \log x$: This term grows faster than $x^3$, but slightly slower than $x^4$, due to the logarithmic factor.

3. Find the dominant term:

   • Between $2x^2$ and $x^3 \log x$, the term $x^3 \log x$ grows faster than $2x^2$ as $x \to \infty$.
   • Therefore, the term $x^3 \log x$ dominates the growth of $f(x)$.

4. Estimate the order of the function:

   • The logarithmic term $\log x$ grows slower than any power of $x$, but $x^3 \log x$ grows slightly faster than $x^3$.
   • Since we are interested in the least integer $n$ such that $f(x) = O(x^n)$, we should approximate the growth of $x^3 \log x$ as $O(x^3)$, because $\log x$ grows very slowly compared to powers of $x$.

Conclusion:

Thus, the least integer $n$ such that $f(x) = O(x^n)$ is $\boxed{3}$.

Would you like further clarification on any step or concept? Here are some related questions to explore:

1. What is Big-O notation, and how is it used to estimate the growth of functions?
2. How does the logarithmic function compare to polynomial functions in terms of growth?
3. Why does the term $x^3 \log x$ dominate over $2x^2$ for large values of $x$?
4. What is the difference between Big-O and little-o notation?
5. How would the result change if $f(x)$ included an exponential term?

Tip: When comparing terms for Big-O, always focus on the highest power of $x$ and consider additional slow-growing factors like logarithms only after comparing the powers.