Step 1: Understanding the Concept:
The integrand consists of the exponential function $e^x$ multiplied by an algebraic expression. This specific structure strongly suggests we should look for the standard integrability form $\int e^x [f(x) + f'(x)] dx$. We need to algebraically manipulate the expression inside the parentheses to identify a function and its exact derivative.
Step 2: Key Formula or Approach:
1. Standard Integral Property: $\int e^x \left[ f(x) + f'(x) \right] dx = e^x f(x) + C$.
2. Distribute terms to expose the $f(x)$ and $f'(x)$ components.
Step 3: Detailed Explanation:
Let the given integral be $I = \int_1^e \frac{e^x}{x}(1 + x \log x) dx$.
First, distribute the factor of $\frac{1}{x}$ into the binomial terms inside the parentheses:
\[ I = \int_1^e e^x \left( \frac{1}{x} + \frac{x \log x}{x} \right) dx \]
Simplify the second term by canceling $x$:
\[ I = \int_1^e e^x \left( \frac{1}{x} + \log x \right) dx \]
Reorder the terms inside the parentheses for clarity:
\[ I = \int_1^e e^x \left( \log x + \frac{1}{x} \right) dx \]
Now, compare this expression with the standard template $\int e^x [f(x) + f'(x)] dx$.
If we select the function to be $f(x) = \log x$, we compute its derivative as $f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$.
The integrand perfectly matches the required pattern: $e^x [f(x) + f'(x)]$.
According to the integration formula, the indefinite integral is $e^x f(x)$:
\[ \int e^x \left( \log x + \frac{1}{x} \right) dx = e^x \log x + C \]
Now, apply the limits $1$ and $e$ for the definite integral:
\[ I = \left[ e^x \log x \right]_1^e \]
Evaluate the expression at the upper limit $e$:
\[ \text{Upper Limit Value} = e^e \log_e(e) = e^e \cdot 1 = e^e \]
Evaluate the expression at the lower limit $1$:
\[ \text{Lower Limit Value} = e^1 \log_e(1) = e \cdot 0 = 0 \]
Subtract the lower limit value from the upper limit value:
\[ I = e^e - 0 = e^e \]
Step 4: Final Answer:
The value of the definite integral is $e^e$.