To solve the integral $\int x^2 e^{2x} \, dx$ using integration by parts, let's break down the steps methodically. Integration by parts follows the formula:

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

Step 1: Choose $u$ and $dv$

We want to simplify the integral by reducing the power of $x^2$, so let's choose:

• $u = x^2$, which implies $du = 2x \, dx$
• $dv = e^{2x} \, dx$, which implies $v = \frac{1}{2} e^{2x}$ (since $\int e^{2x} \, dx = \frac{1}{2} e^{2x}$)

Step 2: Apply the integration by parts formula

Now apply the formula:

$\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x \, dx$

Simplifying:

$= \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx$

Step 3: Apply integration by parts again

Now we need to solve $\int x e^{2x} \, dx$ using integration by parts again. Choose:

• $u = x$, so $du = dx$
• $dv = e^{2x} \, dx$, so $v = \frac{1}{2} e^{2x}$

Apply the formula:

$\int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx$

$= \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}$

Step 4: Substitute back

Now substitute this result into the original equation:

$\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right)$

Simplify the expression:

$\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C$

Where $C$ is the constant of integration.

Final Answer:

$\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C$

Do you have any questions or need further clarification on any of the steps?

Here are five questions for further exploration:

1. How do we determine the best choices for $u$ and $dv$ in integration by parts?
2. What happens if we choose different values for $u$ and $dv$ in this problem?
3. Can we verify the solution by differentiating the final result?
4. What is the difference between using integration by parts once versus multiple times in a problem like this?
5. How does the constant of integration $C$ affect the final answer in indefinite integrals?

Tip: When solving integrals involving both polynomials and exponentials, integration by parts is typically an efficient method to reduce the polynomial term step by step.