The integral given is:

$\int \frac{\sec x}{\sec x + \tan x} \, dx$

To solve this integral, we can proceed by using substitution.

Solution:

1. Substitute $u = \sec x + \tan x$.

2. Differentiate $u$ with respect to $x$: $\frac{du}{dx} = \sec x \tan x + \sec^2 x$ This simplifies to: $du = (\sec x \tan x + \sec^2 x) \, dx$

3. Notice that $\sec x \, dx$ appears in the integral, and we can rewrite: $\sec x \, dx = \frac{du}{\sec x \tan x + \sec^2 x} = \frac{du}{\sec x + \tan x} = \frac{du}{u}$

4. Now the integral becomes: $\int \frac{1}{u} \, du$

5. Integrating this, we get: $\ln |u| + C$

6. Substitute back for $u$: $\ln |\sec x + \tan x| + C$

Final Answer:

$\int \frac{\sec x}{\sec x + \tan x} \, dx = \ln |\sec x + \tan x| + C$

Would you like more details on any step?

Here are some related questions for further exploration:

1. How do we choose substitutions effectively in integrals?
2. Why does $u = \sec x + \tan x$ simplify this integral?
3. Can this method be used for other trigonometric integrals?
4. How would the solution change if the integral was $\int \frac{\sec^2 x}{\sec x + \tan x} \, dx$?
5. What are the applications of integrals involving $\sec x$ and $\tan x$?

Tip: When facing integrals with expressions like $\sec x + \tan x$ in the denominator, consider substitutions based on known derivatives, as it often simplifies the integration process.