`int secx. log (sec x+ tan x ) dx `

Answer

To solve the integral \( I = \int \sec x \cdot \log(\sec x + \tan x) \, dx \), we will use substitution and integration techniques. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = \log(\sec x + \tan x) \).

### Step 2: Differentiate the substitution
Now, we differentiate both sides with respect to \( x \):
\[ dt = \frac{1}{\sec x + \tan x} \cdot \left( \sec x \tan x + \sec^2 x \right) \, dx \]
This simplifies to:
\[ dt = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \, dx \]

### Step 3: Simplify the expression
Notice that:
\[ \sec x + \tan x = \frac{1 + \sin x}{\cos x} \]
Thus, we can rewrite \( dt \) as:
\[ dt = \sec x \, dx \]
This means:
\[ \sec x \, dx = dt \]

### Step 4: Substitute back into the integral
Now we can substitute back into the integral:
\[ I = \int t \, dt \]

### Step 5: Integrate
The integral of \( t \) is:
\[ I = \frac{t^2}{2} + C \]

### Step 6: Substitute \( t \) back
Now we substitute \( t \) back into the expression:
\[ I = \frac{(\log(\sec x + \tan x))^2}{2} + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ I = \frac{(\log(\sec x + \tan x))^2}{2} + C \]