To find the integral $\int x^2 e^{-x} \, dx$, we can use the method of integration by parts. Integration by parts is based on the formula:

$\int u \, dv = uv - \int v \, du$

We need to choose $u$ and $dv$ appropriately. Let's set:

$u = x^2 \quad \text{and} \quad dv = e^{-x} \, dx$

Next, we need to find $du$ and $v$:

$du = 2x \, dx \quad \text{and} \quad v = \int e^{-x} \, dx = -e^{-x}$

Now, apply the integration by parts formula:

$\int x^2 e^{-x} \, dx = uv - \int v \, du$

Substitute $u$, $v$, $du$, and $dv$:

$\int x^2 e^{-x} \, dx = x^2 (-e^{-x}) - \int (-e^{-x}) (2x \, dx)$ $= -x^2 e^{-x} + 2 \int x e^{-x} \, dx$

Now we need to integrate $\int x e^{-x} \, dx$ using integration by parts again. Set:

$u = x \quad \text{and} \quad dv = e^{-x} \, dx$

Then:

$du = dx \quad \text{and} \quad v = -e^{-x}$

Apply the integration by parts formula again:

$\int x e^{-x} \, dx = uv - \int v \, du$

Substitute $u$, $v$, $du$, and $dv$:

$\int x e^{-x} \, dx = x (-e^{-x}) - \int (-e^{-x}) \, dx$ $= -x e^{-x} + \int e^{-x} \, dx$ $= -x e^{-x} - e^{-x}$ $= -e^{-x} (x + 1)$

Now substitute this result back into the original integral:

$\int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 \left( -e^{-x} (x + 1) \right)$ $= -x^2 e^{-x} - 2 e^{-x} (x + 1)$ $= -x^2 e^{-x} - 2x e^{-x} - 2e^{-x}$ $= -e^{-x} (x^2 + 2x + 2)$

Therefore, the integral is:

$\int x^2 e^{-x} \, dx = -e^{-x} (x^2 + 2x + 2) + C$

where $C$ is the constant of integration.

Do you want more details or have any questions?

Here are some related questions for further practice:

1. Evaluate $\int x e^{-x} \, dx$.
2. Evaluate $\int x^3 e^{-x} \, dx$.
3. Find $\int x^2 e^x \, dx$ using integration by parts.
4. Evaluate $\int x^2 \cos(x) \, dx$.
5. Evaluate $\int x^2 \sin(x) \, dx$.
6. Find $\int x^2 e^{2x} \, dx$.
7. Evaluate $\int x^2 \ln(x) \, dx$.
8. Find $\int x^2 e^{-2x} \, dx$.

Tip: When using integration by parts, choose $u$ and $dv$ such that the resulting integral becomes simpler after applying the formula.