To evaluate the integral

$\int (x^2 + 3x) \cos(x) \, dx$

we can use integration by parts, a method useful when the integrand is a product of two functions.

1. Choose $u$ and $dv$: We set: $u = x^2 + 3x \quad \text{and} \quad dv = \cos(x) \, dx$

   Then, calculate $du$ and $v$: $du = (2x + 3) \, dx$ and $v = \int \cos(x) \, dx = \sin(x)$

2. Apply Integration by Parts Formula: The formula for integration by parts is: $\int u \, dv = uv - \int v \, du$

   Applying this to our integral: $\int (x^2 + 3x) \cos(x) \, dx = (x^2 + 3x) \sin(x) - \int \sin(x) (2x + 3) \, dx$

3. **Simplify the Remaining Integral