To evaluate the integral of the expression $-2e^x + e^x \, dx$, we can break it into simpler parts.

The given integral is:

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

Step 1: Break the integral into two separate integrals.

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

Step 2: Integrate each term.

1. The integral of $e^x$ is simply $e^x$, since the derivative of $e^x$ is $e^x$.

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

2. The integral of $-2e^x$ is $-2e^x$, because the constant $-2$ is factored out of the integral.

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

Step 3: Combine the results.

Now, combining the results of the two integrals:

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

where $C$ is the constant of integration.

Thus, the evaluated integral is:

$-2e^x + e^x + C$

Alternatively, you can simplify it to:

$-(e^x) + C$

Would you like further explanation or clarification? Here are some related questions to explore:

1. How do you handle constants when integrating?
2. What happens if the exponent is a linear function of $x$, like $e^{2x}$?
3. Can you evaluate integrals with other types of exponential functions, such as $e^{3x}$ or $e^{x^2}$?
4. How does the process differ if the expression inside the integral has a coefficient that isn't constant?
5. What are some integration rules for combining exponential functions?

Tip: When integrating exponential functions, the basic rule is that the integral of $e^{kx}$ with respect to $x$ is $\frac{e^{kx}}{k}$, where $k$ is a constant.