`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 \( \int a^x \left( \ln x + \ln a \cdot \ln \left( \frac{x}{e} \right)^x \right) dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ \ln x + \ln a \cdot \ln \left( \frac{x}{e} \right)^x \]
Using the properties of logarithms, we can rewrite \( \ln \left( \frac{x}{e} \right)^x \) as:
\[ \ln \left( \frac{x}{e} \right)^x = x \cdot \ln \left( \frac{x}{e} \right) = x \cdot (\ln x - \ln e) = x \cdot (\ln x - 1) \]
Thus, we can rewrite the integrand as:
\[ \ln x + \ln a \cdot x \cdot (\ln x - 1) \]
This simplifies to:
\[ \ln x + \ln a \cdot x \cdot \ln x - \ln a \cdot x \]
Combining terms gives us:
\[ (1 + \ln a \cdot x) \ln x - \ln a \cdot x \]

### Step 2: Set up the integral
Now we can rewrite the integral:
\[ \int a^x \left( (1 + \ln a \cdot x) \ln x - \ln a \cdot x \right) dx \]
This can be split into two separate integrals:
\[ \int a^x (1 + \ln a \cdot x) \ln x \, dx - \ln a \int a^x x \, dx \]

### Step 3: Solve the first integral using integration by parts
Let:
- \( u = \ln x \) and \( dv = a^x (1 + \ln a \cdot x) dx \)
Then we differentiate and integrate:
- \( du = \frac{1}{x} dx \)
- \( v = \frac{a^x}{\ln a} (1 + \ln a \cdot x) \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We can compute:
\[ \int \ln x \cdot a^x (1 + \ln a \cdot x) dx = \frac{\ln x \cdot a^x}{\ln a} (1 + \ln a \cdot x) - \int \frac{a^x}{\ln a} (1 + \ln a \cdot x) \cdot \frac{1}{x} dx \]

### Step 4: Solve the second integral
The second integral \( \int a^x x \, dx \) can also be solved using integration by parts:
Let:
- \( u = x \) and \( dv = a^x dx \)
Then:
- \( du = dx \)
- \( v = \frac{a^x}{\ln a} \)

Thus:
\[ \int x a^x dx = x \cdot \frac{a^x}{\ln a} - \int \frac{a^x}{\ln a} dx \]
The integral \( \int a^x dx = \frac{a^x}{\ln a} \).

### Step 5: Combine results
After solving both integrals, we combine the results:
\[ \int a^x \left( (1 + \ln a \cdot x) \ln x - \ln a \cdot x \right) dx = \text{(result from first integral)} - \ln a \cdot \left( x \cdot \frac{a^x}{\ln a} - \frac{a^x}{(\ln a)^2} \right) + C \]

### Final Result
The final result will be:
\[ \int a^x \left( \ln x + \ln a \cdot \ln \left( \frac{x}{e} \right)^x \right) dx = \text{(combined result)} + C \]