`int_(1)^(3)|(2-x)log_(e )x|dx` is equal to:

Options

A. `(3)/(2)log_(e )3+(1)/(2)`
B. `"log"_(e )(16)/(3sqrt(2))(1)/(2)`
C. `-(3)/(2)log_(e )3-(1)/(2)`
D. None of these

Answer

To solve the integral \( \int_{1}^{3} |(2-x) \log_e x| \, dx \), we first need to analyze the expression inside the absolute value, \( (2-x) \log_e x \).

### Step 1: Determine the intervals for the absolute value
The expression \( (2-x) \) changes sign at \( x = 2 \). Therefore, we need to evaluate the integral in two parts:
1. From \( x = 1 \) to \( x = 2 \) where \( (2-x) \) is positive.
2. From \( x = 2 \) to \( x = 3 \) where \( (2-x) \) is negative.

Thus, we can rewrite the integral as:
\[ \int_{1}^{3} |(2-x) \log_e x| \, dx = \int_{1}^{2} (2-x) \log_e x \, dx + \int_{2}^{3} -(2-x) \log_e x \, dx \]

### Step 2: Evaluate the first integral \( \int_{1}^{2} (2-x) \log_e x \, dx \)
Using integration by parts, let:
- \( u = \log_e x \) ⇒ \( du = \frac{1}{x} \, dx \)
- \( dv = (2-x) \, dx \) ⇒ \( v = 2x - \frac{x^2}{2} \)

Now applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Substituting the values:
\[ \int (2-x) \log_e x \, dx = \left( \log_e x \left( 2x - \frac{x^2}{2} \right) \right) \bigg|_{1}^{2} - \int (2x - \frac{x^2}{2}) \cdot \frac{1}{x} \, dx \]

Calculating the boundary term:
\[ = \left( \log_e 2 \left( 2(2) - \frac{2^2}{2} \right) - \log_e 1 \left( 2(1) - \frac{1^2}{2} \right) \right) \]
\[ = \left( \log_e 2 \cdot 2 - 0 \right) = 2 \log_e 2 \]

Now, we need to evaluate the integral:
\[ \int (2 - \frac{x}{2}) \, dx = \int 2 \, dx - \frac{1}{2} \int x \, dx = 2x - \frac{x^2}{4} \bigg|_{1}^{2} \]
Calculating this:
\[ = \left( 2(2) - \frac{2^2}{4} \right) - \left( 2(1) - \frac{1^2}{4} \right) = (4 - 1) - (2 - \frac{1}{4}) = 3 - 1.75 = 1.25 \]

Thus,
\[ \int_{1}^{2} (2-x) \log_e x \, dx = 2 \log_e 2 - 1.25 \]

### Step 3: Evaluate the second integral \( \int_{2}^{3} -(2-x) \log_e x \, dx \)
This can be rewritten as:
\[ \int_{2}^{3} (x-2) \log_e x \, dx \]
Using integration by parts again:
Let:
- \( u = \log_e x \) ⇒ \( du = \frac{1}{x} \, dx \)
- \( dv = (x-2) \, dx \) ⇒ \( v = \frac{x^2}{2} - 2x \)

Now applying integration by parts:
\[ \int (x-2) \log_e x \, dx = \left( \log_e x \left( \frac{x^2}{2} - 2x \right) \right) \bigg|_{2}^{3} - \int \left( \frac{x^2}{2} - 2x \right) \cdot \frac{1}{x} \, dx \]

Calculating the boundary term:
\[ = \left( \log_e 3 \left( \frac{3^2}{2} - 2(3) \right) - \log_e 2 \left( \frac{2^2}{2} - 2(2) \right) \right) \]
\[ = \left( \log_e 3 \cdot \left( \frac{9}{2} - 6 \right) - \log_e 2 \cdot (2 - 4) \right) = \left( \log_e 3 \cdot \left( \frac{-3}{2} \right) + 2 \log_e 2 \right) \]

Now, we need to evaluate the integral:
\[ \int \left( \frac{x^2}{2} - 2x \right) \cdot \frac{1}{x} \, dx = \int \left( \frac{x}{2} - 2 \right) \, dx = \frac{x^2}{4} - 2x \bigg|_{2}^{3} \]
Calculating this:
\[ = \left( \frac{3^2}{4} - 2(3) \right) - \left( \frac{2^2}{4} - 2(2) \right) = \left( \frac{9}{4} - 6 \right) - \left( 1 - 4 \right) = \left( \frac{9}{4} - \frac{24}{4} \right) - (-3) = \left( -\frac{15}{4} + 3 \right) = \frac{3}{4} \]

Thus,
\[ \int_{2}^{3} (x-2) \log_e x \, dx = \left( -\frac{3}{2} \log_e 3 + 2 \log_e 2 \right) - \frac{3}{4} \]

### Step 4: Combine the results
Now we can combine both parts:
\[ \int_{1}^{3} |(2-x) \log_e x| \, dx = \left( 2 \log_e 2 - 1.25 \right) + \left( -\frac{3}{2} \log_e 3 + 2 \log_e 2 - \frac{3}{4} \right) \]
Combine like terms:
\[ = 4 \log_e 2 - \frac{3}{2} \log_e 3 - 1.25 - \frac{3}{4} \]
Calculating the constants:
\[ = 4 \log_e 2 - \frac{3}{2} \log_e 3 - \frac{5}{4} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{1}^{3} |(2-x) \log_e x| \, dx = 4 \log_e 2 - \frac{3}{2} \log_e 3 - \frac{5}{4} \]