Integration by Parts: ∫x²eˣ dx

Setup and Method

Integration by Parts Formula: ∫u dv = uv − ∫v du

This integral requires repeated application because differentiating x² takes two steps to reach a constant.

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First Application of Integration by Parts

Choose:

Variable	Assignment	Reason
u = x²	du = 2x dx	Differentiates toward 0
dv = eˣ dx	v = eˣ	eˣ is its own integral

Applying the formula:

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

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

The remaining integral ∫xeˣ dx still needs integration by parts.

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Second Application of Integration by Parts

Working on ∫xeˣ dx — Choose:

Variable	Assignment
u = x	du = dx
dv = eˣ dx	v = eˣ

Applying the formula:

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

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Combining the Results

Substituting back into the first result:

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

$$= x^2 e^x - 2xe^x + 2e^x$$

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

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Verification by Differentiation

Differentiate F(x) = eˣ(x² − 2x + 2) using the product rule:

$$F'(x) = \frac{d}{dx}[e^x] \cdot (x^2-2x+2) + e^x \cdot \frac{d}{dx}[x^2-2x+2]$$

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

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

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

$$\boxed{F'(x) = x^2 e^x} \checkmark$$

This matches our original integrand, confirming the answer is correct.

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Key Pattern (LIATE Rule)

When choosing u, pick the function that simplifies when differentiated: Logarithms → Inverse trig → Algebraic (polynomials) → Trig → Exponentials

For xⁿeˣ, always let u = xⁿ and dv = eˣ dx, applying integration by parts n times until the polynomial disappears.

To solve $\int x^2 e^x , dx$, we use integration by parts twice. The formula is:

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

First Application of Integration by Parts

Choice: We want to reduce the polynomial power, so let:

• $u = x^2$  → $du = 2x , dx$
• $dv = e^x , dx$  → $v = e^x$

Applying the formula: $$\int x^2 e^x \, dx = x^2 \cdot e^x - \int e^x \cdot 2x \, dx$$

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

Second Application of Integration by Parts

Now we evaluate $\int x e^x , dx$:

Choice:

• $u = x$  → $du = dx$
• $dv = e^x , dx$  → $v = e^x$

Applying the formula: $$\int x e^x \, dx = x \cdot e^x - \int e^x \, dx$$

$$= x e^x - e^x + C_1$$

$$= e^x(x - 1) + C_1$$

Combining the Results

Substitute back into our first result:

$$\int x^2 e^x \, dx = x^2 e^x - 2[e^x(x - 1)] + C$$

$$= x^2 e^x - 2x e^x + 2e^x + C$$

Factor out $e^x$:

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

Verification by Differentiation

Let $F(x) = e^x(x^2 - 2x + 2)$. Using the product rule $[f \cdot g]' = f'g + fg'$:

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

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

$$= e^x[(x^2 - 2x + 2) + (2x - 2)]$$

$$= e^x[x^2 - 2x + 2 + 2x - 2]$$

$$= e^x \cdot x^2$$

$$= x^2 e^x \quad \checkmark$$

The derivative matches the original integrand, confirming our solution is correct.