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

Options

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

Answer

To solve the integral \( \int_{\frac{1}{e}}^{e} |\log x| \, dx \), we first need to analyze the behavior of the function \( |\log x| \) over the interval from \( \frac{1}{e} \) to \( e \).

### Step 1: Determine the intervals where \( \log x \) is positive or negative
- For \( x < 1 \), \( \log x < 0 \) (negative).
- For \( x = 1 \), \( \log x = 0 \).
- For \( x > 1 \), \( \log x > 0 \) (positive).

Given the limits of integration \( \frac{1}{e} \) and \( e \):
- \( \log \left( \frac{1}{e} \right) = -1 \) (negative)
- \( \log(1) = 0 \) (zero)
- \( \log(e) = 1 \) (positive)

### Step 2: Break the integral into two parts
We can split the integral at \( x = 1 \):
\[ \int_{\frac{1}{e}}^{e} |\log x| \, dx = \int_{\frac{1}{e}}^{1} -\log x \, dx + \int_{1}^{e} \log x \, dx \]

### Step 3: Calculate the first integral \( \int_{\frac{1}{e}}^{1} -\log x \, dx \)
Using integration by parts, let:
- \( u = -\log x \) → \( du = -\frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int -\log x \, dx = -x \log x + \int x \cdot \frac{1}{x} \, dx = -x \log x + x \]

Now we evaluate from \( \frac{1}{e} \) to \( 1 \):
\[ \left[-x \log x + x\right]_{\frac{1}{e}}^{1} = \left[-1 \cdot \log(1) + 1\right] - \left[-\frac{1}{e} \cdot \log\left(\frac{1}{e}\right) + \frac{1}{e}\right] \]
Calculating this gives:
\[ = [0 + 1] - \left[-\frac{1}{e} \cdot (-1) + \frac{1}{e}\right] = 1 - \left[\frac{1}{e} + \frac{1}{e}\right] = 1 - \frac{2}{e} \]

### Step 4: Calculate the second integral \( \int_{1}^{e} \log x \, dx \)
Using integration by parts again:
Let:
- \( u = \log x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

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

Now evaluate from \( 1 \) to \( e \):
\[ \left[x \log x - x\right]_{1}^{e} = \left[e \cdot \log e - e\right] - \left[1 \cdot \log 1 - 1\right] \]
Calculating this gives:
\[ = [e \cdot 1 - e] - [0 - 1] = [e - e] + 1 = 1 \]

### Step 5: Combine both integrals
Now we combine the results:
\[ \int_{\frac{1}{e}}^{e} |\log x| \, dx = \left(1 - \frac{2}{e}\right) + 1 = 2 - \frac{2}{e} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{\frac{1}{e}}^{e} |\log x| \, dx = 2 \left(1 - \frac{1}{e}\right) \]