`int_(1)^(e) log (x) dx=`

Options

A. 1
B. e
C. `e-1`
D. `1-e`

Answer

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

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

### Step-by-Step Solution:

**Step 1: Choose \(u\) and \(dv\)**
Let:
- \(u = \log(x)\) (which is our first function)
- \(dv = dx\) (which is our second function)

**Step 2: Differentiate \(u\) and integrate \(dv\)**
Now we differentiate \(u\) and integrate \(dv\):
- \(du = \frac{1}{x} \, dx\)
- \(v = x\)

**Step 3: Apply the integration by parts formula**
Using the integration by parts formula:

\[ \int \log(x) \, dx = x \log(x) - \int x \cdot \frac{1}{x} \, dx \]

This simplifies to:

\[ \int \log(x) \, dx = x \log(x) - \int 1 \, dx \]

**Step 4: Evaluate the remaining integral**
The integral of 1 is simply \(x\):

\[ \int \log(x) \, dx = x \log(x) - x + C \]

**Step 5: Evaluate the definite integral from 1 to \(e\)**
Now we need to evaluate this from 1 to \(e\):

\[ \int_{1}^{e} \log(x) \, dx = \left[ x \log(x) - x \right]_{1}^{e} \]

**Step 6: Calculate the upper limit \(e\)**
Substituting \(x = e\):

\[ e \log(e) - e = e \cdot 1 - e = e - e = 0 \]

**Step 7: Calculate the lower limit \(1\)**
Substituting \(x = 1\):

\[ 1 \log(1) - 1 = 1 \cdot 0 - 1 = -1 \]

**Step 8: Combine the results**
Now we combine the results from the upper and lower limits:

\[ \int_{1}^{e} \log(x) \, dx = 0 - (-1) = 1 \]

### Final Answer:
The value of the integral \(\int_{1}^{e} \log(x) \, dx\) is \(1\).

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