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

Options

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

Answer

To solve the integral

\[ \int_{e^{-1}}^{e^{2}} \left| \frac{\log_e x}{x} \right| dx, \]

we first need to analyze the function \(\frac{\log_e x}{x}\) to determine where it is positive and negative.

### Step 1: Identify the turning point
The function \(\frac{\log_e x}{x}\) changes its sign at \(x = 1\) because \(\log_e 1 = 0\). Therefore, we will split the integral at this point.

### Step 2: Split the integral
We can write the integral as:

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

### Step 3: Evaluate the first integral
For the first integral, we have:

\[ \int_{e^{-1}}^{1} -\frac{\log_e x}{x} dx. \]

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

Thus, the integral becomes:

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

### Step 4: Evaluate the second integral
For the second integral, we have:

\[ \int_{1}^{e^{2}} \frac{\log_e x}{x} dx. \]

Using the same substitution \(t = \log_e x\), the limits change as follows:
- When \(x = 1\), \(t = 0\).
- When \(x = e^{2}\), \(t = 2\).

Thus, the integral becomes:

\[ \int_{0}^{2} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - 0 = \frac{4}{2} = 2. \]

### Step 5: Combine the results
Now, we combine the results of both integrals:

\[ \int_{e^{-1}}^{e^{2}} \left| \frac{\log_e x}{x} \right| dx = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}. \]

Thus, the value of the integral is

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