The integral in the image is:

$\int \frac{-1}{e^x} \, dx$

We can solve this step by step:

Step 1: Simplify the integrand

The term $\frac{-1}{e^x}$ can be rewritten as:

$\frac{-1}{e^x} = -e^{-x}$

Thus, the integral becomes:

$\int -e^{-x} \, dx$

Step 2: Integrate $e^{-x}$

The integral of $e^{-x}$ with respect to $x$ is:

$\int e^{-x} \, dx = -e^{-x} + C$

where $C$ is the constant of integration.

Step 3: Include the negative sign

Adding the negative sign outside the integral, we get:

$\int -e^{-x} \, dx = -\left( -e^{-x} \right) + C = e^{-x} + C$

Final Answer:

$\int \frac{-1}{e^x} \, dx = e^{-x} + C$

Do you want further explanations on any step?

Related Questions:

1. What is the integral of $e^x$ and $e^{-x}$ with respect to $x$?
2. How do you compute definite integrals involving exponential functions?
3. What are some properties of exponential functions and their integrals?
4. How do you handle integrals with negative exponents?
5. Can you explain how to integrate exponential functions multiplied by polynomials?

Tip:

Always check for simplifications in integrals involving exponential functions before integrating. Simplifying first makes calculations easier!