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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.

Evaluate the integral \( \int \frac{1}{e^x + e^{-x}} \, dx \)

This solution provides a step-by-step evaluation of the integral \( \int \frac{1}{e^x + e^{-x}} \, dx \). Learn about substitution techniques, simplifying exponential expressions, and applying the inverse trigonometric function formula. Suitable for college-level calculus students.

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