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To evaluate the integral

\[
\int x^2 e^x \, dx
\]

we can use the technique of integration by parts multiple times. Let’s proceed with the steps:

### Step 1: Integration by Parts (First Application)
For integration by parts, we use the formula:

\[
\int u \, dv = uv - \int v \, du
\]

Choose:
- \( u = x^2 \) (which means \( du = 2x \, dx \))
- \( dv = e^x \, dx \) (which gives \( v = e^x \))

Then,
\[
\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx
\]

### Step 2: Integration by Parts (Second Application)
Now, we need to evaluate the integral \(\int 2x e^x \, dx\).

Choose:
- \( u = 2x \) (which means \( du = 2 \, dx \))
- \( dv = e^x \, dx \) (which gives \( v = e^x \))

Then,
\[
\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx
\]

### Step 3: Final Integration
Now we integrate \(\int 2 e^x \, dx\), which is straightforward:

\[
\int 2 e^x \, dx = 2 e^x
\]

### Putting it All Together
Substituting back, we get:

\[
\int x^2 e^x \, dx = x^2 e^x - (2x e^x - 2 e^x)
\]

Simplifying, we find:

\[
\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2 e^x + C
\]

where \( C \) is the constant of integration.

---

Would you like more details, or do you have any questions?

### Related Questions
1. How does integration by parts work, and when is it useful?
2. What are alternative methods for evaluating integrals like this one?
3. How would the answer change if the integral was \( \int x^n e^x \, dx \) for any integer \( n \)?
4. Could substitution help simplify similar integrals?
5. How can we confirm the result by differentiating the answer?

### Tip
When using integration by parts repeatedly, keep track of terms carefully to avoid errors in substitution.

Evaluate the integral \( \int x^2 e^x \, dx \)

This guide provides a step-by-step solution for evaluating the integral \( \int x^2 e^x \, dx \) using integration by parts. Explore how to handle integrals involving polynomial and exponential functions, with multiple applications of integration by parts.

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