The value of the integral `int _(e ^(-1))^(e ^(2))|(ln x )/(x)|dx ` is:

Options

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

Answer

To solve the integral \( \int_{e^{-1}}^{e^{2}} \left| \frac{\ln x}{x} \right| dx \), we can follow these steps:

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

First, we need to analyze the function \( \frac{\ln x}{x} \) to determine where it is positive or negative in the interval \( [e^{-1}, e^{2}] \).

- At \( x = e^{-1} \):
\[ \frac{\ln(e^{-1})}{e^{-1}} = \frac{-1}{\frac{1}{e}} = -e \]
- At \( x = 1 \):
\[ \frac{\ln(1)}{1} = 0 \]
- At \( x = e^{2} \):
\[ \frac{\ln(e^{2})}{e^{2}} = \frac{2}{e^{2}} \]

The function \( \frac{\ln x}{x} \) is negative for \( x \in (e^{-1}, 1) \) and positive for \( x \in (1, e^{2}) \).

### Step 2: Rewrite the integral using the absolute value

Since \( \frac{\ln x}{x} \) is negative from \( e^{-1} \) to \( 1 \) and positive from \( 1 \) to \( e^{2} \), we can express the integral as:

\[ \int_{e^{-1}}^{e^{2}} \left| \frac{\ln x}{x} \right| dx = -\int_{e^{-1}}^{1} \frac{\ln x}{x} dx + \int_{1}^{e^{2}} \frac{\ln x}{x} dx \]

### Step 3: Change of variables

Let \( t = \ln x \), then \( dx = e^t dt \) and \( x = e^t \). The limits change as follows:
- When \( x = e^{-1} \), \( t = -1 \)
- When \( x = 1 \), \( t = 0 \)
- When \( x = e^{2} \), \( t = 2 \)

Now, substituting these into the integral gives:

\[ -\int_{-1}^{0} t \, dt + \int_{0}^{2} t \, dt \]

### Step 4: Evaluate the integrals

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

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

### Step 5: Combine the results

Now combine the results from the two integrals:
\[ \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]

### Final Answer

Thus, the value of the integral is:
\[ \int_{e^{-1}}^{e^{2}} \left| \frac{\ln x}{x} \right| dx = \frac{5}{2} \]