To solve

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

we use integration by parts twice.

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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:

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

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Step 2: Second integration by parts

Now solve:

$$\int x e^x dx$$

Again let:

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

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

Then:

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

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

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

Thus:

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

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Step 3: Substitute back

From Step 1:

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

Expand if desired:

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

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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Verification 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 $e^x$:

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

Simplify inside:

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

$$= x^2 e^x$$

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

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

Here is the step-by-step solution to evaluate the integral, followed by the verification.

Part 1: Solving the Integral

To solve $\int x^2 e^x , dx$, we will use Integration by Parts. The formula for integration by parts is: $$\int u \, dv = uv - \int v \, du$$

Step 1: Choose $u$ and $dv$ for the first integration by parts. Using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to choose $u$, we select the algebraic term $x^2$. Let:

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

Step 2: Apply the integration by parts formula. $$\int x^2 e^x \, dx = x^2 e^x - \int e^x (2x) \, dx$$ $$\int x^2 e^x \, dx = x^2 e^x - 2 \int x e^x \, dx$$

Step 3: Apply integration by parts a second time. We now need to evaluate the new integral: $\int x e^x , dx$. Let:

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

Apply the formula to this part: $$\int x e^x \, dx = x e^x - \int e^x \, dx$$ $$\int x e^x \, dx = x e^x - e^x$$

Step 4: Substitute this back into the equation from Step 2. $$\int x^2 e^x \, dx = x^2 e^x - 2 \left( x e^x - e^x \right) + C$$

Step 5: Distribute and simplify. $$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C$$

To make it look cleaner, we can factor out $e^x$: Final Answer: $$e^x(x^2 - 2x + 2) + C$$

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Part 2: Verifying by Differentiating

By the Fundamental Theorem of Calculus, if we differentiate our answer, we should get the original integrand ($x^2 e^x$) back.

We need to find: $$\frac{d}{dx} \left[ e^x(x^2 - 2x + 2) + C \right]$$

Step 1: Apply the Product Rule. The product rule states that $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$. Let:

• $f(x) = e^x \implies f'(x) = e^x$
• $g(x) = x^2 - 2x + 2 \implies g'(x) = 2x - 2$

(Note: The derivative of the constant $C$ is $0$, so it drops out).

Step 2: Write out the derivative. $$\frac{d}{dx} \left[ e^x(x^2 - 2x + 2) \right] = (e^x)(x^2 - 2x + 2) + (e^x)(2x - 2)$$

Step 3: Factor out the $e^x$ and simplify. $$= e^x \left[ (x^2 - 2x + 2) + (2x - 2) \right]$$

Now, combine the like terms inside the brackets:

• The $-2x$ and $+2x$ cancel out.
• The $+2$ and $-2$ cancel out.

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

Because the derivative matches our original integrand, our integration is verified to be 100% correct!