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To solve the given exercise using integration by parts, let's break down the problem step by step. The goal is to prove a specific integral and then use the result to deduce another expression.

### Problem:
You are asked to prove:
\[
\int x^n e^{-\beta x} dx = -\frac{x^n}{\beta} e^{-\beta x} + \frac{n}{\beta} \int x^{n-1} e^{-\beta x} dx
\]
and then use this result to deduce:
\[
\int x^2 e^{-\beta x} dx = \frac{1}{\beta^3} e^{-\beta x} \left( \beta^2 x^2 - 2 \beta x + 2 \right) e^{-\beta x} + C
\]

### Step 1: Apply Integration by Parts
Let’s start with proving the first integral using integration by parts. Recall the integration by parts formula:
\[
\int u dv = uv - \int v du
\]
Here, let's choose:
- \( u = x^n \) (so \( du = n x^{n-1} dx \))
- \( dv = e^{-\beta x} dx \) (so \( v = -\frac{1}{\beta} e^{-\beta x} \))

Using the integration by parts formula:
\[
\int x^n e^{-\beta x} dx = -\frac{1}{\beta} x^n e^{-\beta x} + \frac{n}{\beta} \int x^{n-1} e^{-\beta x} dx
\]
This proves the required formula.

### Step 2: Deduce the Specific Integral
Now, using the result from Step 1 for \( n = 2 \), we get:
\[
\int x^2 e^{-\beta x} dx = -\frac{x^2}{\beta} e^{-\beta x} + \frac{2}{\beta} \int x e^{-\beta x} dx
\]

Next, apply the same integration by parts to \( \int x e^{-\beta x} dx \) with:
- \( u = x \) and \( dv = e^{-\beta x} dx \)
- \( du = dx \) and \( v = -\frac{1}{\beta} e^{-\beta x} \)

So,
\[
\int x e^{-\beta x} dx = -\frac{x}{\beta} e^{-\beta x} + \frac{1}{\beta} \int e^{-\beta x} dx
\]
The integral \( \int e^{-\beta x} dx \) is straightforward:
\[
\int e^{-\beta x} dx = -\frac{1}{\beta} e^{-\beta x} + C
\]

Substituting back, we get:
\[
\int x e^{-\beta x} dx = -\frac{x}{\beta} e^{-\beta x} - \frac{1}{\beta^2} e^{-\beta x} + C
\]

Finally, substitute this into the earlier equation:
\[
\int x^2 e^{-\beta x} dx = -\frac{x^2}{\beta} e^{-\beta x} + \frac{2}{\beta} \left(-\frac{x}{\beta} e^{-\beta x} - \frac{1}{\beta^2} e^{-\beta x}\right)
\]
Simplifying this expression:
\[
\int x^2 e^{-\beta x} dx = \left(-\frac{x^2}{\beta} + \frac{2x}{\beta^2} - \frac{2}{\beta^3}\right) e^{-\beta x} + C
\]

Thus, the deduced integral is:
\[
\int x^2 e^{-\beta x} dx = \frac{1}{\beta^3} e^{-\beta x} \left( \beta^2 x^2 - 2 \beta x + 2 \right) + C
\]

This completes the proof.

Would you like any further details or explanations on specific steps?

Here are five additional related questions for practice:
1. How would the approach differ if \( \beta \) were negative?
2. What would be the result if \( n = 1 \) in the first integral?
3. How do the results change if we use a different method, like substitution, instead of integration by parts?
4. How can integration by parts be extended to integrals with higher powers of \( x \)?
5. Can you derive a similar formula for \( \int x^n e^{\beta x} dx \)?

**Tip:** Remember that careful selection of \( u \) and \( dv \) in integration by parts can significantly simplify the problem.

Learn how to prove and deduce integrals using integration by parts with a detailed step-by-step solution. This problem involves proving a specific integral and deducing another expression using the integral calculus method of integration by parts. Suitable for advanced college students and anyone studying integral calculus.

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