To solve the integral

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

we can simplify the expression by rewriting the integrand. Notice that:

$e^x + e^{-x} = \frac{e^{2x} + 1}{e^x}.$

Thus, we can rewrite the integral as:

$\int \frac{1}{e^x + e^{-x}} \, dx = \int \frac{e^x}{e^{2x} + 1} \, dx.$

Now, let's make a substitution to simplify further. Let:

$u = e^x \Rightarrow du = e^x \, dx \Rightarrow dx = \frac{du}{e^x} = \frac{du}{u}.$

Substituting $e^x = u$ into the integral, we get:

$\int \frac{e^x}{e^{2x} + 1} \, dx = \int \frac{u}{u^2 + 1} \cdot \frac{du}{u} = \int \frac{1}{u^2 + 1} \, du.$

This integral is now in a standard form:

$\int \frac{1}{u^2 + 1} \, du = \arctan(u) + C.$

Finally, substitute back $u = e^x$ to get the answer:

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

The solution to the integral is:

$\boxed{\arctan(e^x) + C}.$

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

Here are some related questions for further exploration:

1. How would you integrate $\int \frac{1}{e^x - e^{-x}} \, dx$?
2. What is the derivative of $\arctan(e^x)$?
3. Can you evaluate $\int \frac{x}{e^x + e^{-x}} \, dx$?
4. How do you simplify expressions involving hyperbolic functions like $\cosh(x)$ and $\sinh(x)$?
5. What are the properties of inverse trigonometric functions in integration?

Tip: When you encounter an integrand with exponential terms like $e^x + e^{-x}$, consider using substitutions or rewriting in terms of hyperbolic functions to simplify.