`inte^(xloga).e^(x)dx` is equal to

Options

A. `(ae)^(x)`
B. `((ae)^(x))/(log(ae))`
C. `(e^(x))/(1+loga)`
D. none of these

Answer

To solve the integral \( \int e^{x \log a} \cdot e^x \, dx \), we can follow these steps:

### Step 1: Combine the Exponents
We start with the integral:
\[ \int e^{x \log a} \cdot e^x \, dx \]
Using the property of exponents, we can combine the terms:
\[ e^{x \log a} \cdot e^x = e^{x \log a + x} = e^{x (\log a + 1)} \]
Thus, the integral can be rewritten as:
\[ \int e^{x (\log a + 1)} \, dx \]

### Step 2: Use the Integration Formula
We know that the integral of \( e^{kx} \) is given by:
\[ \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \]
In our case, \( k = \log a + 1 \). Therefore, we can apply this formula:
\[ \int e^{x (\log a + 1)} \, dx = \frac{1}{\log a + 1} e^{x (\log a + 1)} + C \]

### Step 3: Substitute Back
Now, we can substitute back to express the result in terms of \( a \):
\[ = \frac{1}{\log a + 1} e^{x (\log a + 1)} + C \]
This can be simplified to:
\[ = \frac{e^{x} \cdot e^{x \log a}}{\log a + 1} + C \]
Since \( e^{x \log a} = a^x \), we can rewrite it as:
\[ = \frac{a^x e^x}{\log a + 1} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int e^{x \log a} \cdot e^x \, dx = \frac{a^x e^x}{\log a + 1} + C \]