To solve the integral of $2e^{2x} - 2$, we'll proceed term by term.

Integral:

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

Step 1: Integral of $2e^{2x}$

We use substitution for the term $2e^{2x}$.

Let $u = 2x$, so $du = 2dx$, which gives us: $\int 2e^{2x} \, dx = e^{2x} + C_1$

Step 2: Integral of $-2$

The integral of $-2$ with respect to $x$ is: $\int -2 \, dx = -2x + C_2$

Final Result:

Combining both integrals, we get: $\int (2e^{2x} - 2) \, dx = e^{2x} - 2x + C$ where $C$ is the constant of integration (since $C_1$ and $C_2$ can be combined into a single constant).

Would you like further details or clarifications on this process? Here are five related questions:

1. How do you apply substitution in more complex integrals?
2. What is the general rule for integrating exponential functions?
3. How do constants affect integration?
4. Can you demonstrate how to differentiate the result to verify correctness?
5. How do you handle definite integrals of similar expressions?

Tip: Always check your answer by differentiating the result—it should give you the original integrand.