`int_(1)^(e) (e^(x) (1+xlog x))/(x) dx`

Answer

To solve the integral

\[ I = \int_{1}^{e} \frac{e^{x}(1+x\log{x})}{x} \, dx, \]

we can start by separating the terms inside the integral:

\[ I = \int_{1}^{e} e^{x} \left( \frac{1}{x} + \log{x} \right) \, dx. \]

This can be rewritten as:

\[ I = \int_{1}^{e} \frac{e^{x}}{x} \, dx + \int_{1}^{e} e^{x} \log{x} \, dx. \]

Now we will evaluate each integral separately.

### Step 1: Evaluate \(\int_{1}^{e} \frac{e^{x}}{x} \, dx\)

We will use integration by parts for this integral. Let:

- \( u = \log{x} \) \(\Rightarrow du = \frac{1}{x} \, dx\)
- \( dv = e^{x} \, dx \) \(\Rightarrow v = e^{x}\)

Now applying integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ \int \frac{e^{x}}{x} \, dx = \log{x} \cdot e^{x} - \int e^{x} \cdot \frac{1}{x} \, dx. \]

Evaluating from 1 to \(e\):

\[ \left[ \log{x} \cdot e^{x} \right]_{1}^{e} - \int_{1}^{e} e^{x} \cdot \frac{1}{x} \, dx. \]

### Step 2: Evaluate \(\int_{1}^{e} e^{x} \log{x} \, dx\)

We will again use integration by parts. Let:

- \( u = \log{x} \) \(\Rightarrow du = \frac{1}{x} \, dx\)
- \( dv = e^{x} \, dx \) \(\Rightarrow v = e^{x}\)

So,

\[ \int e^{x} \log{x} \, dx = \log{x} \cdot e^{x} - \int e^{x} \cdot \frac{1}{x} \, dx. \]

### Step 3: Combine the results

Now we can combine the results from both integrals:

\[ I = \left[ e^{x} \log{x} \right]_{1}^{e} + \int_{1}^{e} e^{x} \log{x} \, dx. \]

Substituting the limits:

\[ I = \left[ e^{e} \cdot 1 - e^{1} \cdot 0 \right] + \int_{1}^{e} e^{x} \log{x} \, dx. \]

### Step 4: Final evaluation

Now we can see that both integrals will cancel each other out:

\[ I = e^{e} - 0 + 0 = e^{e}. \]

So the final answer is:

\[ \int_{1}^{e} \frac{e^{x}(1+x\log{x})}{x} \, dx = e^{e}. \]