To evaluate the integral $\int (2x + 3) \, dx$, we integrate each term separately.

1. For $2x$, the integral is $\int 2x \, dx = x^2$ (since $\int x \, dx = \frac{x^2}{2}$, and multiplying by 2 gives $x^2$).
2. For $3$, the integral is $\int 3 \, dx = 3x$ (since the integral of a constant is the constant multiplied by $x$).

So, the integral becomes:

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

Where $C$ is the constant of integration.

Summary:

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

Would you like more details or have any questions on this?

Here are some related questions for further practice:

1. How do you evaluate $\int (ax + b) \, dx$, where $a$ and $b$ are constants?
2. What happens when you integrate a higher degree polynomial, like $\int (3x^3 + 2x^2 + x) \, dx$?
3. Can you explain how integration of trigonometric functions differs from polynomials?
4. How would you evaluate an indefinite integral with a negative exponent, like $\int x^{-2} \, dx$?
5. What is the difference between definite and indefinite integrals?

Tip: Always remember to add the constant $C$ when performing indefinite integration, as it represents the family of all antiderivatives.