(i) `int sec x dx=|log| tanx|+c`
(ii) `int sec dx=log |secx+tanx|+c`

Answer

To solve the integrals given in the question, we will derive the integration of sec x step by step.

### Step-by-Step Solution

1. **Start with the integral of sec x:**
\[ \int \sec x \, dx \]

2. **Multiply the integrand by a clever form of 1:**
We can multiply the numerator and the denominator by \(\sec x + \tan x\):
\[ \int \sec x \, dx = \int \frac{\sec x (\sec x + \tan x)}{\sec x + \tan x} \, dx \]

3. **Rewrite the integral:**
This gives us:
\[ \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx \]

4. **Substitute \(u = \sec x + \tan x\):**
Now, let:
\[ u = \sec x + \tan x \]
Then, we need to find \(du\):
\[ du = (\sec x \tan x + \sec^2 x) \, dx \]

5. **Express dx in terms of du:**
Rearranging gives:
\[ dx = \frac{du}{\sec x \tan x + \sec^2 x} \]

6. **Substituting back into the integral:**
The integral now becomes:
\[ \int \frac{du}{u} \]

7. **Integrate:**
The integral of \(\frac{1}{u}\) is:
\[ \log |u| + C \]

8. **Substituting back for u:**
Replace \(u\) with \(\sec x + \tan x\):
\[ \int \sec x \, dx = \log |\sec x + \tan x| + C \]

### Conclusion
Thus, the correct integral is:
\[ \int \sec x \, dx = \log |\sec x + \tan x| + C \]

### Verification of the Given Options
- (i) \(\int \sec x \, dx = | \log | \tan x | + C\) is **false**.
- (ii) \(\int \sec x \, dx = \log |\sec x + \tan x| + C\) is **true**.