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The integral you provided is:

\[
\int_0^\infty x^4 e^{-x} x^{n-1} \, dx
\]

Simplifying the integrand:

\[
\int_0^\infty x^{n+3} e^{-x} \, dx
\]

This is a form of the **Gamma function**. The Gamma function is defined as:

\[
\Gamma(s) = \int_0^\infty x^{s-1} e^{-x} \, dx
\]

Here, \( x^{n+3} \) suggests that \( s = n+4 \), so the integral can be expressed as:

\[
\int_0^\infty x^{n+3} e^{-x} \, dx = \Gamma(n+4)
\]

Thus, the result of the integral is:

\[
\Gamma(n+4)
\]

Let me know if you would like further details or clarifications.

Here are 5 related questions:
1. What is the Gamma function and its general properties?
2. How does the Gamma function relate to factorials?
3. Can the Gamma function be generalized to complex numbers?
4. What are some applications of the Gamma function in physics?
5. How do we evaluate integrals involving exponentials and powers of \(x\) in general?

**Tip:** The Gamma function generalizes the factorial function, where \(\Gamma(n) = (n-1)!\).

Evaluate the integral from 0 to infinity of x^4 * e^(-x) * x^(n-1) dx.

Discover how to solve the integral ∫ x^4 * e^(-x) * x^(n-1) dx using the Gamma function. This solution demonstrates the connection between integrals of exponential functions and the Gamma function, suitable for advanced calculus and mathematical physics students.

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