ChatGPT vs Perplexity for Math

We need to compute

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

using integration by parts, and then check by differentiating.

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Step 1: Use integration by parts

Recall:

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

Choose:

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

Then:

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

So now we need:

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

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Step 2: Integrate $\int x e^x\,dx$ by parts again

Again use integration by parts:

Choose:

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

Then:

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

$$= x e^x - e^x$$

$$= e^x(x-1)$$

So:

$$2\int x e^x\,dx = 2e^x(x-1)$$

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Step 3: Substitute back

From Step 1:

$$\int x^2 e^x\,dx = x^2 e^x - 2e^x(x-1)$$

Factor out $e^x$:

$$= e^x\bigl(x^2 - 2x + 2\bigr) + C$$

So the answer is:

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

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Step 4: Verify by differentiation

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:

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

$$= x^2 e^x$$

This matches the original integrand, so the result is correct.

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Final Answer

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

Verified since

$$\frac{d}{dx}\left[e^x(x^2 - 2x + 2)\right] = x^2 e^x$$

The indefinite integral is $\int x^2 e^x \, dx = e^x (x^2 - 2x + 2) + C$[1][2][4].

Step-by-Step Solution Using Integration by Parts

Integration by parts uses the formula $\int u \, dv = uv - \int v \, du$. Apply LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) to choose $u$ (algebraic: $x^2$) and $dv$ (exponential: $e^x dx$) first[1][2].

1. Let $u = x^2$, so $du = 2x \, dx$; let $dv = e^x \, dx$, so $v = e^x$[1][2][4].
   $$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx$$

2. For $\int 2x e^x \, dx$, apply parts again: let $u = 2x$, so $du = 2 \, dx$; $dv = e^x \, dx$, so $v = e^x$[1][2][4].
   $$\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx = 2x e^x - 2 e^x + C_1$$

3. Substitute back:
   $$\int x^2 e^x \, dx = x^2 e^x - (2x e^x - 2 e^x) + C = e^x (x^2 - 2x + 2) + C$$[4]

Verification by Differentiation

Differentiate $e^x (x^2 - 2x + 2)$ using the product rule:
$$\frac{d}{dx} [e^x (x^2 - 2x + 2)] = e^x (x^2 - 2x + 2) + e^x (2x - 2) = e^x (x^2 - 2x + 2 + 2x - 2) = x^2 e^x$$ This matches the integrand, confirming the result[1][7].