The integral given in the image is:

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

This can be solved using integration by parts, where we choose one function to differentiate and another to integrate. Let's define:

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

Using the integration by parts formula:

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

Substituting the values:

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

We still need to compute $\int 2x e^x \, dx$. Once again, we'll use integration by parts, letting:

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

Applying integration by parts again:

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

Now, integrating $\int 2 e^x \, dx$ gives:

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

Thus, we have:

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

Substituting this result back into the original equation:

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

Simplifying:

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

where $C$ is the constant of integration.

Let me know if you'd like further details on any step! Here are some related questions:

1. How is integration by parts derived from the product rule?
2. What are the common strategies for solving integrals involving polynomials and exponentials?
3. Can you apply integration by parts when functions contain trigonometric terms?
4. How does integration by parts differ from substitution?
5. How do we determine the best choice for $u$ and $dv$ in integration by parts?

Tip: When using integration by parts, a good strategy is to choose $u$ such that its derivative simplifies the integral!