`int_1^elogxdx=`

Options

A. 1
B. `-1`
C. `+1`
D. 0

Answer

To solve the integral \( \int_1^e \log x \, dx \), we will use integration by parts. The integration by parts formula is given by:

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

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \log x \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

### 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
Now we apply 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: Solve the Integral
Now, we compute the integral:

\[ \int 1 \, dx = x \]

Thus, we have:

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

### Step 5: Evaluate the Definite Integral
Now we need to evaluate the definite integral from 1 to \( e \):

\[ \int_1^e \log x \, dx = \left[ x \log x - x \right]_1^e \]

### Step 6: Substitute the Limits
Now we will substitute the limits:

1. Substitute \( x = e \):
\[ e \log e - e = e \cdot 1 - e = e - e = 0 \]

2. Substitute \( x = 1 \):
\[ 1 \log 1 - 1 = 1 \cdot 0 - 1 = 0 - 1 = -1 \]

### Step 7: Calculate the Final Result
Now we combine the results from the limits:

\[ \int_1^e \log x \, dx = 0 - (-1) = 1 \]

Thus, the final answer is:

\[ \int_1^e \log x \, dx = 1 \]