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

Answer

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

### Step 1: Substitution
Let:
\[ t = \log(\sec x + \tan x) \]
Then, we need to find \( dt \).

### Step 2: Differentiate \( t \)
Using the chain rule, we differentiate \( t \):
\[ dt = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x) \, dx \]
Now, we need to differentiate \( \sec x + \tan x \).

### Step 3: Differentiate \( \sec x + \tan x \)
The derivative of \( \sec x \) is \( \sec x \tan x \) and the derivative of \( \tan x \) is \( \sec^2 x \):
\[ \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \]
Thus,
\[ dt = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \, dx \]

### Step 4: Factor out \( \sec x \)
We can factor \( \sec x \) from the numerator:
\[ dt = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \, dx \]
This simplifies to:
\[ dt = \sec x \, dx \]

### Step 5: Substitute back into the integral
Now we can rewrite the integral:
\[ \int \sec x \log(\sec x + \tan x) \, dx = \int t \, dt \]

### Step 6: Integrate
The integral of \( t \) is:
\[ \int t \, dt = \frac{t^2}{2} + C \]

### Step 7: Substitute back for \( t \)
Now substitute back \( t = \log(\sec x + \tan x) \):
\[ \frac{(\log(\sec x + \tan x))^2}{2} + C \]

### Final Answer
Thus, the final result is:
\[ \int \sec x \log(\sec x + \tan x) \, dx = \frac{(\log(\sec x + \tan x))^2}{2} + C \]

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