The value of the integral `int_(e^(-1))^(e^(2)) |(log_(e)x)/(x)|dx` is

Options

A. `(3)/(2)`
B. `(5)/(2)`
C. 3
D. 5

Explanation

To solve the integral \( I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx \), we will break it down into manageable steps.

### Step 1: Analyze the Function
The function we are integrating is \( f(x) = \left| \frac{\log_e x}{x} \right| \). We need to determine where this function is positive or negative within the limits of integration.

### Step 2: Identify Critical Points
The logarithm function \( \log_e x \) is zero at \( x = 1 \). It is negative for \( 0 < x < 1 \) and positive for \( x > 1 \). Thus:
- For \( x \in [e^{-1}, 1) \), \( \log_e x < 0 \) and hence \( f(x) = -\frac{\log_e x}{x} \).
- For \( x \in (1, e^2] \), \( \log_e x > 0 \) and hence \( f(x) = \frac{\log_e x}{x} \).

### Step 3: Split the Integral
We can split the integral at \( x = 1 \):
\[ I = \int_{e^{-1}}^{1} -\frac{\log_e x}{x} \, dx + \int_{1}^{e^2} \frac{\log_e x}{x} \, dx \]

### Step 4: Change of Variables
For both integrals, we can use the substitution \( t = \log_e x \), which gives \( dt = \frac{1}{x} dx \) or \( dx = x dt = e^t dt \). The limits change as follows:
- When \( x = e^{-1} \), \( t = -1 \)
- When \( x = 1 \), \( t = 0 \)
- When \( x = e^2 \), \( t = 2 \)

Thus, we rewrite the integrals:
1. For the first integral:
\[ \int_{e^{-1}}^{1} -\frac{\log_e x}{x} \, dx = \int_{-1}^{0} -t \, dt \]
2. For the second integral:
\[ \int_{1}^{e^2} \frac{\log_e x}{x} \, dx = \int_{0}^{2} t \, dt \]

### Step 5: Evaluate the Integrals
Now we evaluate each integral:
1. For the first integral:
\[ \int_{-1}^{0} -t \, dt = -\left[ \frac{t^2}{2} \right]_{-1}^{0} = -\left( 0 - \frac{(-1)^2}{2} \right) = \frac{1}{2} \]
2. For the second integral:
\[ \int_{0}^{2} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - 0 = 2 \]

### Step 6: Combine the Results
Now we combine the results of the two integrals:
\[ I = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{5}{2}} \]