To evaluate the integral of $(3x + 4)^2 \, dx$, we will expand the integrand first and then integrate term by term.

Step 1: Expand the expression

$(3x + 4)^2 = (3x + 4)(3x + 4)$ $= 9x^2 + 12x \cdot 4 + 16$ $= 9x^2 + 24x + 16$

Step 2: Integrate term by term

Now, we can integrate each term with respect to $x$: $\int (9x^2 + 24x + 16) \, dx$

This can be broken down into three separate integrals: $\int 9x^2 \, dx + \int 24x \, dx + \int 16 \, dx$

Step 3: Compute each integral

1. $\int 9x^2 \, dx$:

$9 \cdot \int x^2 \, dx = 9 \cdot \frac{x^3}{3} = 3x^3$

2. $\int 24x \, dx$:

$24 \cdot \int x \, dx = 24 \cdot \frac{x^2}{2} = 12x^2$

3. $\int 16 \, dx$:

$16 \cdot \int 1 \, dx = 16x$

Step 4: Combine the results

Now, add the results together and include the constant of integration $C$: $\int (3x + 4)^2 \, dx = 3x^3 + 12x^2 + 16x + C$

This is the final answer.

Would you like further details or have any questions?

Here are 5 related questions:

1. How would you integrate $(5x - 7)^2 \, dx$?
2. What is the derivative of the result we just obtained?
3. How do you solve integrals involving trigonometric functions?
4. How do you approach integrating by parts?
5. What is the integral of a product of polynomials?

Tip: Always expand the integrand when it's in a simple polynomial form before integrating, as this often simplifies the process.