To solve the integral $\int x^2 e^{3x} \, dx$, we can apply the integration by parts method multiple times.

The formula for integration by parts is:

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

Step 1: First Integration by Parts

We let:

• $u = x^2$, hence $du = 2x \, dx$
• $dv = e^{3x} \, dx$, hence $v = \frac{1}{3} e^{3x}$

Now, apply the integration by parts formula:

$\int x^2 e^{3x} \, dx = \frac{1}{3} x^2 e^{3x} - \int \frac{1}{3} \cdot 2x e^{3x} \, dx$ $= \frac{1}{3} x^2 e^{3x} - \frac{2}{3} \int x e^{3x} \, dx$

Step 2: Second Integration by Parts

Now, apply integration by parts again on $\int x e^{3x} \, dx$.

Let:

• $u = x$, hence $du = dx$
• $dv = e^{3x} \, dx$, hence $v = \frac{1}{3} e^{3x}$

Now apply the formula again:

$\int x e^{3x} \, dx = \frac{1}{3} x e^{3x} - \int \frac{1}{3} e^{3x} \, dx$ $= \frac{1}{3} x e^{3x} - \frac{1}{9} e^{3x}$

Step 3: Combine Results

Now substitute back into the previous equation:

$\int x^2 e^{3x} \, dx = \frac{1}{3} x^2 e^{3x} - \frac{2}{3} \left( \frac{1}{3} x e^{3x} - \frac{1}{9} e^{3x} \right)$ $= \frac{1}{3} x^2 e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27} e^{3x}$

Final Answer:

Thus, the integral is:

$\int x^2 e^{3x} \, dx = \frac{1}{3} x^2 e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27} e^{3x} + C$

where $C$ is the constant of integration.

Would you like more details on any specific part of the solution?

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Here are 5 related questions you might find useful:

1. How does integration by parts work in general?
2. Can you apply integration by parts to solve $\int x^3 e^{3x} \, dx$?
3. What are the common techniques for solving integrals involving polynomials and exponentials?
4. What other methods can be used to solve integrals of this type?
5. How do we choose the parts (i.e., $u$ and $dv$) in integration by parts?

Tip: When applying integration by parts multiple times, keep track of the terms carefully to avoid errors.