`int e^(x log a) e^(x) dx ` is equal to

Options

A. `(a^(x))/(log ae)+c`
B. `(e^(x))/(1+log_(e)a)+c`
C. `(ae)^(x)+c`
D. `((ae)^(x))/(log_(e)ae)+c`

Explanation

To solve the integral \( \int e^{(x \log a)} e^{x} \, dx \), we can simplify the expression first.

### Step-by-Step Solution:

1. **Combine the Exponents**:
We can combine the exponents in the integrand:
\[ e^{(x \log a)} e^{x} = e^{(x \log a + x)} = e^{x(\log a + 1)} \]

2. **Rewrite the Integral**:
Now we can rewrite the integral:
\[ \int e^{x(\log a + 1)} \, dx \]

3. **Use the Exponential Integral Formula**:
The integral of \( e^{kx} \) is given by:
\[ \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \]
where \( k = \log a + 1 \).

4. **Apply the Formula**:
Applying this formula to our integral:
\[ \int e^{x(\log a + 1)} \, dx = \frac{1}{\log a + 1} e^{x(\log a + 1)} + C \]

5. **Final Expression**:
We can express the final answer as:
\[ \frac{e^{x(\log a + 1)}}{\log a + 1} + C \]

### Final Answer:
\[ \int e^{(x \log a)} e^{x} \, dx = \frac{e^{x(\log a + 1)}}{\log a + 1} + C \]

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