We want to compute

$$\int x^2 e^x \, dx$$

using integration by parts.

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Step 1: First integration by parts

Recall:

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

Choose:

$$u = x^2 \quad \Rightarrow \quad du = 2x\,dx$$

$$dv = e^x\,dx \quad \Rightarrow \quad v = e^x$$

Apply the formula:

$$\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx$$

So now we need to compute:

$$\int 2x e^x\,dx = 2\int x e^x\,dx$$

Thus,

$$\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx$$

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Step 2: Second integration by parts

Now evaluate:

$$\int x e^x\,dx$$

Choose:

$$u = x \quad \Rightarrow \quad du = dx$$

$$dv = e^x\,dx \quad \Rightarrow \quad v = e^x$$

Apply integration by parts:

$$\int x e^x\,dx = x e^x - \int e^x\,dx$$

$$\int x e^x\,dx = x e^x - e^x$$

So,

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

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Step 3: Substitute back

From Step 1:

$$\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx$$

Substitute the result:

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

Distribute the $-2$:

$$\int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x$$

Factor out $e^x$:

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

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Final Answer

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

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Step 4: Verify by differentiating

Differentiate:

$$F(x) = e^x(x^2 - 2x + 2)$$

Use the product rule:

$$F'(x) = e^x(x^2 - 2x + 2) + e^x(2x - 2)$$

Factor out $e^x$:

$$F'(x) = e^x\big[(x^2 - 2x + 2) + (2x - 2)\big]$$

Simplify inside:

$$(x^2 - 2x + 2) + (2x - 2) = x^2$$

So:

$$F'(x) = e^x x^2$$

$$F'(x) = x^2 e^x$$

which matches the integrand.

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So the result is verified:

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

Alright, let’s go step-by-step.

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1. Problem

$$\int x^2 e^x \, dx$$

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2. Integration by parts

We recall the formula: $$\int u \, dv = uv - \int v \, du$$ Let’s choose: $$u = x^2, \quad dv = e^x \, dx$$ Then: $$du = 2x \, dx, \quad v = e^x$$

So: $$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx$$ $$= x^2 e^x - 2 \int x e^x \, dx$$

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3. Integrate $\int x e^x \, dx$ by parts

Let $u = x,\ dv = e^x dx$, then $du = dx,\ v = e^x$: $$\int x e^x \, dx = x e^x - \int e^x \, dx$$ $$= x e^x - e^x + C_1$$

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4. Substitute back

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

Factor $e^x$: $$\int x^2 e^x \, dx = e^x \left( x^2 - 2x + 2 \right) + C$$

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5. Verification by differentiation

Let $F(x) = e^x (x^2 - 2x + 2)$.

Using product rule: $$F'(x) = e^x (x^2 - 2x + 2) + e^x (2x - 2)$$ Factor $e^x$: $$F'(x) = e^x \left[ x^2 - 2x + 2 + 2x - 2 \right]$$ Simplify inside: $$x^2 - 2x + 2 + 2x - 2 = x^2$$ So: $$F'(x) = e^x \cdot x^2$$ which is the integrand $x^2 e^x$. Correct.

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$$\boxed{e^x(x^2 - 2x + 2) + C}$$