What is `int_(1)^(e) xln x dx` equal to ?

Options

A. `(e + 1)/(4)`
B. `(e^(2) + 1)/(4)`
C. `(e-1)/(4)`
D. `(e^(2) - 1)/(4)`

Explanation

To solve the integral \( I = \int_{1}^{e} x \ln x \, dx \), we will use integration by parts. The formula for integration by parts is:

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

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = \ln x \) (which implies \( du = \frac{1}{x} \, dx \))
- \( dv = x \, dx \) (which implies \( v = \frac{x^2}{2} \))

### Step 2: Apply the integration by parts formula

Using the integration by parts formula, we have:

\[ I = \left[ \ln x \cdot \frac{x^2}{2} \right]_{1}^{e} - \int_{1}^{e} \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]

### Step 3: Simplify the integral

The integral simplifies to:

\[ I = \left[ \ln x \cdot \frac{x^2}{2} \right]_{1}^{e} - \int_{1}^{e} \frac{x}{2} \, dx \]

### Step 4: Evaluate the boundary term

Now, we evaluate the boundary term:

\[ \left[ \ln x \cdot \frac{x^2}{2} \right]_{1}^{e} = \left( \ln e \cdot \frac{e^2}{2} \right) - \left( \ln 1 \cdot \frac{1^2}{2} \right) \]

Since \( \ln e = 1 \) and \( \ln 1 = 0 \), we get:

\[ = \frac{e^2}{2} - 0 = \frac{e^2}{2} \]

### Step 5: Evaluate the remaining integral

Next, we evaluate the remaining integral:

\[ \int_{1}^{e} \frac{x}{2} \, dx = \frac{1}{2} \int_{1}^{e} x \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{e} \]

Calculating this gives:

\[ = \frac{1}{2} \left( \frac{e^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{e^2}{2} - \frac{1}{2} \right) = \frac{1}{4}(e^2 - 1) \]

### Step 6: Combine the results

Now, substituting back into our expression for \( I \):

\[ I = \frac{e^2}{2} - \frac{1}{4}(e^2 - 1) \]

### Step 7: Simplify

Now, let's simplify this:

\[ I = \frac{e^2}{2} - \frac{1}{4}e^2 + \frac{1}{4} = \left( \frac{2}{4}e^2 - \frac{1}{4}e^2 \right) + \frac{1}{4} = \frac{1}{4}e^2 + \frac{1}{4} \]

Combining the terms:

\[ I = \frac{1}{4}(e^2 + 1) \]

### Final Answer

Thus, the value of the integral \( \int_{1}^{e} x \ln x \, dx \) is:

\[ \boxed{\frac{1}{4}(e^2 + 1)} \]