What is `int_1^2 logx dx` equal to

Options

A. `log e^2`
B. 1
C. `log_e(4/e)`
D. `log_e (e/4)`

Answer

To solve the integral \( \int_1^2 \log x \, dx \), 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:
- \( u = \log x \) (which is a logarithmic function)
- \( dv = dx \) (which is a simple differential)

Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = x \]

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

Substituting \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula, we have:

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

This simplifies to:

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

### Step 3: Evaluate the integral

Now, we can evaluate the integral:

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

Thus, we have:

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

### Step 4: Evaluate the definite integral from 1 to 2

Now we need to evaluate this from 1 to 2:

\[ \int_1^2 \log x \, dx = \left[ x \log x - x \right]_1^2 \]

Calculating the upper limit (when \( x = 2 \)):

\[ 2 \log 2 - 2 \]

Calculating the lower limit (when \( x = 1 \)):

\[ 1 \log 1 - 1 = 0 - 1 = -1 \]

### Step 5: Combine the results

Now, we combine the results:

\[ \int_1^2 \log x \, dx = \left(2 \log 2 - 2\right) - (-1) \]

This simplifies to:

\[ \int_1^2 \log x \, dx = 2 \log 2 - 2 + 1 = 2 \log 2 - 1 \]

### Final Result

Thus, the value of the definite integral \( \int_1^2 \log x \, dx \) is:

\[ \int_1^2 \log x \, dx = 2 \log 2 - 1 \]