` int sec x tan x sqrt(sec^(2) x+1)dx`

Answer

To solve the integral \( \int \sec x \tan x \sqrt{\sec^2 x + 1} \, dx \), we will use the substitution method. Here are the steps to solve the integral:

### Step 1: Substitution
Let \( \sec x = t \). Then, the derivative of \( \sec x \) is:
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \quad \Rightarrow \quad \sec x \tan x \, dx = dt \]
Thus, we can rewrite the integral in terms of \( t \):
\[ \int \sec x \tan x \sqrt{\sec^2 x + 1} \, dx = \int \sqrt{t^2 + 1} \, dt \]

### Step 2: Simplifying the Integral
Now, we need to evaluate the integral:
\[ \int \sqrt{t^2 + 1} \, dt \]
This integral can be solved using the formula:
\[ \int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln |x + \sqrt{x^2 + a^2}| + C \]
In our case, \( a = 1 \). Therefore, we have:
\[ \int \sqrt{t^2 + 1} \, dt = \frac{t}{2} \sqrt{t^2 + 1} + \frac{1}{2} \ln |t + \sqrt{t^2 + 1}| + C \]

### Step 3: Back Substitution
Now, we substitute back \( t = \sec x \):
\[ = \frac{\sec x}{2} \sqrt{\sec^2 x + 1} + \frac{1}{2} \ln |\sec x + \sqrt{\sec^2 x + 1}| + C \]

### Step 4: Simplifying the Expression
Since \( \sqrt{\sec^2 x + 1} = \sqrt{(\sec^2 x + \tan^2 x)} = \sqrt{1 + \tan^2 x} = \sec x \), we can simplify further:
\[ = \frac{\sec x}{2} \cdot \sec x + \frac{1}{2} \ln |\sec x + \sqrt{\sec^2 x + 1}| + C \]
\[ = \frac{\sec^2 x}{2} + \frac{1}{2} \ln |\sec x + \sec x| + C \]
\[ = \frac{\sec^2 x}{2} + \frac{1}{2} \ln |2 \sec x| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sec x \tan x \sqrt{\sec^2 x + 1} \, dx = \frac{\sec^2 x}{2} + \frac{1}{2} \ln |2 \sec x| + C \]