If `int e^(sec x)(sec x tan x f(x)+(sec x tan x + sec^(2) x))dx = e^(sec x)f(x) + C`, then a possible choice of f(x) is

Options

A. `sec x + x tan x - (1)/(2)`
B. `x sec x + tan x + (1)/(2)`
C. `sec x - tan x - (1)/(2)`
D. `sec x + tan x + (1)/(2)`

Explanation

To solve the given problem, we need to find a possible choice of \( f(x) \) given the equation:

\[ \int e^{\sec x} \left( \sec x \tan x f(x) + \sec x \tan x + \sec^2 x \right) dx = e^{\sec x} f(x) + C \]

### Step-by-Step Solution:

1. **Differentiate Both Sides with Respect to \( x \)**:
We start by differentiating both sides of the equation with respect to \( x \). This is based on the principle that differentiation and integration are inverse operations.

\[ \frac{d}{dx} \left( \int e^{\sec x} \left( \sec x \tan x f(x) + \sec x \tan x + \sec^2 x \right) dx \right) = \frac{d}{dx} \left( e^{\sec x} f(x) + C \right) \]

The left-hand side simplifies to:

\[ e^{\sec x} \left( \sec x \tan x f(x) + \sec x \tan x + \sec^2 x \right) \]

The right-hand side uses the product rule:

\[ \frac{d}{dx} \left( e^{\sec x} f(x) \right) = e^{\sec x} \sec x \tan x f(x) + e^{\sec x} f'(x) \]

Thus, we have:

\[ e^{\sec x} \left( \sec x \tan x f(x) + \sec x \tan x + \sec^2 x \right) = e^{\sec x} \sec x \tan x f(x) + e^{\sec x} f'(x) \]

2. **Cancel \( e^{\sec x} \)**:
Since \( e^{\sec x} \) is never zero, we can safely divide both sides by \( e^{\sec x} \):

\[ \sec x \tan x f(x) + \sec x \tan x + \sec^2 x = \sec x \tan x f(x) + f'(x) \]

3. **Rearranging the Equation**:
We can rearrange the equation to isolate \( f'(x) \):

\[ \sec x \tan x + \sec^2 x = f'(x) \]

4. **Integrate Both Sides**:
To find \( f(x) \), we integrate both sides with respect to \( x \):

\[ f(x) = \int \left( \sec x \tan x + \sec^2 x \right) dx \]

The integral of \( \sec x \tan x \) is \( \sec x \), and the integral of \( \sec^2 x \) is \( \tan x \):

\[ f(x) = \sec x + \tan x + C \]

### Final Answer:
Thus, a possible choice of \( f(x) \) is:

\[ f(x) = \sec x + \tan x + C \]