`int e^(x log a ) e^(x) dx` is equal to A) `(a^(x))/( log ae) + C` B) `( e^(x))/( 1+log a )`+ C C) `( ae )^(x) +C` D) `((ae)^(x))/( log ae) +C`

Options

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

Answer

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

### Step 1: Combine the Exponents
Since both terms in the integral have the same base \( e \), we can combine the exponents:
\[ e^{x \log a} e^{x} = e^{x \log a + x} = e^{x (\log a + 1)} \]
Thus, the integral becomes:
\[ \int e^{x (\log a + 1)} \, dx \]

### Step 2: Use the Exponential Integral Formula
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 substitute back the expression for \( e^{x (\log a + 1)} \):
\[ = \frac{1}{\log a + 1} e^{x} e^{x \log a} + C = \frac{1}{\log a + 1} e^{x (1 + \log a)} + C \]
Since \( e^{x (1 + \log a)} = (ae)^{x} \), we can rewrite the integral as:
\[ = \frac{(ae)^{x}}{\log a + 1} + C \]

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