`int ((1+tanx))/((x+log secx))dx`

Answer

To solve the integral \( I = \int \frac{1 + \tan x}{x + \log(\sec x)} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Identify the substitution
Let’s set \( t = x + \log(\sec x) \). We will differentiate \( t \) with respect to \( x \).

**Hint:** Choose a substitution that simplifies the integral, especially the denominator.

### Step 2: Differentiate \( t \)
To differentiate \( t \), we need to find \( \frac{dt}{dx} \):
\[ \frac{dt}{dx} = 1 + \frac{d}{dx}(\log(\sec x)) \]
Using the chain rule, we have:
\[ \frac{d}{dx}(\log(\sec x)) = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x \]
Thus,
\[ \frac{dt}{dx} = 1 + \tan x \]
This implies:
\[ dt = (1 + \tan x) \, dx \]

**Hint:** Differentiate your substitution to relate \( dt \) and \( dx \).

### Step 3: Rewrite the integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int \frac{1 + \tan x}{t} \, dx \]
Substituting \( dt = (1 + \tan x) \, dx \) gives us:
\[ I = \int \frac{1}{t} \, dt \]

**Hint:** Substitute \( dx \) in terms of \( dt \) using your earlier result.

### Step 4: Integrate
The integral \( \int \frac{1}{t} \, dt \) is a standard integral:
\[ I = \log |t| + C \]

**Hint:** Remember the integral of \( \frac{1}{t} \) is \( \log |t| \).

### Step 5: Substitute back
Now, substitute back \( t = x + \log(\sec x) \):
\[ I = \log |x + \log(\sec x)| + C \]

**Hint:** Always substitute back to the original variable to express your final answer.

### Final Answer
Thus, the integral is:
\[ \int \frac{1 + \tan x}{x + \log(\sec x)} \, dx = \log |x + \log(\sec x)| + C \]