`inte^x[log (sec x +tan x)+sec x]dx = `

Options

A. `e^x log tan x + c`
B. `e^x log (sec x + tan x)+c`
C. `e^x log sec x+c`
D. None of these

Answer

To solve the integral \( \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx \), we can break it down into two parts:

1. **Separate the integral**:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = \int e^x \log(\sec x + \tan x) \, dx + \int e^x \sec x \, dx \]

2. **Focus on the second integral**:
We will use integration by parts for the integral \( \int e^x \sec x \, dx \). Recall the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can choose:
- \( u = \sec x \) (which we will differentiate)
- \( dv = e^x dx \) (which we will integrate)

3. **Differentiate and integrate**:
- Differentiate \( u \):
\[ du = \sec x \tan x \, dx \]
- Integrate \( dv \):
\[ v = e^x \]

4. **Apply integration by parts**:
Now, substituting into the integration by parts formula:
\[ \int e^x \sec x \, dx = e^x \sec x - \int e^x \sec x \tan x \, dx \]

5. **Combine the results**:
Now, we can substitute back into our original integral:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = \int e^x \log(\sec x + \tan x) \, dx + \left( e^x \sec x - \int e^x \sec x \tan x \, dx \right) \]

6. **Notice the cancellation**:
The integral \( \int e^x \sec x \tan x \, dx \) appears in both the positive and negative terms, allowing us to cancel them:
\[ \int e^x \log(\sec x + \tan x) \, dx + e^x \sec x \]

7. **Final result**:
Thus, we have:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = e^x \log(\sec x + \tan x) + C \]

### Final Answer:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = e^x \log(\sec x + \tan x) + C \]