What is `int_(e^(-1))^(e^(2))` |(ln x)/(x)|dx` equal to ?

Options

A. `3/2`
B. `5/2`
C. `3`
D. `4`

Explanation

To solve the integral \( I = \int_{e^{-1}}^{e^{2}} \left| \frac{\ln x}{x} \right| dx \), we need to analyze the expression inside the absolute value and determine where it is positive or negative.

### Step 1: Determine the sign of \( \frac{\ln x}{x} \)

1. **Find the critical points**:
- The function \( \ln x \) is zero at \( x = 1 \).
- For \( x < 1 \), \( \ln x < 0 \) and hence \( \frac{\ln x}{x} < 0 \).
- For \( x = 1 \), \( \frac{\ln x}{x} = 0 \).
- For \( x > 1 \), \( \ln x > 0 \) and hence \( \frac{\ln x}{x} > 0 \).

### Step 2: Break the integral into intervals

Given the limits of integration \( e^{-1} \) and \( e^{2} \), we know:
- \( e^{-1} < 1 < e^{2} \)

Thus, we can split the integral at \( x = 1 \):

\[ I = \int_{e^{-1}}^{1} -\frac{\ln x}{x} \, dx + \int_{1}^{e^{2}} \frac{\ln x}{x} \, dx \]

### Step 3: Evaluate the first integral

For the first integral:

\[ \int_{e^{-1}}^{1} -\frac{\ln x}{x} \, dx \]

Using the substitution \( t = \ln x \), then \( dt = \frac{1}{x} dx \) or \( dx = e^t dt \). The limits change as follows:
- When \( x = e^{-1} \), \( t = -1 \)
- When \( x = 1 \), \( t = 0 \)

Thus, the integral becomes:

\[ \int_{-1}^{0} -t \, dt = \left[ -\frac{t^2}{2} \right]_{-1}^{0} = 0 - \left( -\frac{(-1)^2}{2} \right) = \frac{1}{2} \]

### Step 4: Evaluate the second integral

For the second integral:

\[ \int_{1}^{e^{2}} \frac{\ln x}{x} \, dx \]

Using the same substitution \( t = \ln x \):
- When \( x = 1 \), \( t = 0 \)
- When \( x = e^{2} \), \( t = 2 \)

Thus, the integral becomes:

\[ \int_{0}^{2} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - 0 = 2 \]

### Step 5: Combine the results

Now we combine both results:

\[ I = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{5}{2}} \]