`inta^(x) (ln x + ln a. ln ((x)/(e))^(x))dx=`

Options

A. `a ^(x) ln((e)/(x ))^(2x) +C`
B. `a ^(x) ln ((x)/(e ))^(x) +C`
C. `a ^(x)+ ln ((x)/(e ))^(x) +C`
D. None of these

Answer

To solve the integral \( I = \int a^x \left( \ln x + \ln a \cdot \ln \left( \frac{x}{e} \right) \right) dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ \ln a \cdot \ln \left( \frac{x}{e} \right) = \ln a \cdot (\ln x - \ln e) = \ln a \cdot \ln x - \ln a \]
Thus, we can rewrite the integral as:
\[ I = \int a^x \left( \ln x + \ln a \cdot \ln x - \ln a \right) dx \]
This simplifies to:
\[ I = \int a^x \left( (1 + \ln a) \ln x - \ln a \right) dx \]

### Step 2: Separate the integral
We can separate the integral into two parts:
\[ I = (1 + \ln a) \int a^x \ln x \, dx - \ln a \int a^x \, dx \]

### Step 3: Solve the first integral using integration by parts
Let \( u = \ln x \) and \( dv = a^x dx \).
Then, we have:
\[ du = \frac{1}{x} dx \quad \text{and} \quad v = \frac{a^x}{\ln a} \]
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int \ln x \cdot a^x \, dx = \frac{a^x \ln x}{\ln a} - \int \frac{a^x}{\ln a} \cdot \frac{1}{x} \, dx \]

### Step 4: Solve the second integral
The second integral \( \int a^x \, dx \) is straightforward:
\[ \int a^x \, dx = \frac{a^x}{\ln a} \]

### Step 5: Combine results
Now we can substitute back into our expression for \( I \):
\[ I = (1 + \ln a) \left( \frac{a^x \ln x}{\ln a} - \int \frac{a^x}{\ln a} \cdot \frac{1}{x} \, dx \right) - \ln a \cdot \frac{a^x}{\ln a} \]
This simplifies to:
\[ I = \frac{(1 + \ln a) a^x \ln x}{\ln a} - \frac{(1 + \ln a) \cdot a^x}{\ln a} - \frac{a^x}{\ln a} \]

### Step 6: Final simplification
Combining terms, we have:
\[ I = \frac{(1 + \ln a) a^x \ln x - (1 + 2 \ln a) a^x}{\ln a} \]
This can be further simplified if necessary, but we can express it in terms of \( C \) (the constant of integration).

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{(1 + \ln a) a^x \ln x - (1 + 2 \ln a) a^x}{\ln a} + C \]