The value of `int_(1//e )^(e )(|log x|)/(x^(2))dx`, is

Options

A. 2
B. `(2)/(e )`
C. `2(1-(1)/(e ))`
D. 0

Answer

To solve the integral \( I = \int_{\frac{1}{e}}^{e} \frac{|\log x|}{x^2} \, dx \), we will break it down into two parts based on the behavior of the logarithmic function.

### Step 1: Analyze the absolute value of the logarithm
We know that:
- For \( x \in \left(0, 1\right) \), \( \log x < 0 \) (hence \( |\log x| = -\log x \)).
- For \( x \in \left(1, \infty\right) \), \( \log x > 0 \) (hence \( |\log x| = \log x \)).

Given the limits of integration \( \frac{1}{e} \) to \( e \), we can split the integral at \( x = 1 \):
\[ I = \int_{\frac{1}{e}}^{1} \frac{-\log x}{x^2} \, dx + \int_{1}^{e} \frac{\log x}{x^2} \, dx \]

### Step 2: Evaluate the first integral
For the first integral, we have:
\[ I_1 = \int_{\frac{1}{e}}^{1} \frac{-\log x}{x^2} \, dx \]
Substituting \( t = \log x \), we get \( dt = \frac{1}{x} dx \) or \( dx = e^t dt \). The limits change as follows:
- When \( x = \frac{1}{e} \), \( t = \log\left(\frac{1}{e}\right) = -1 \).
- When \( x = 1 \), \( t = \log(1) = 0 \).

Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_{-1}^{0} \frac{-t}{e^{2t}} e^t \, dt = \int_{-1}^{0} -t e^{-t} \, dt \]

### Step 3: Evaluate the second integral
For the second integral:
\[ I_2 = \int_{1}^{e} \frac{\log x}{x^2} \, dx \]
Using the same substitution \( t = \log x \), we have:
- When \( x = 1 \), \( t = 0 \).
- When \( x = e \), \( t = 1 \).

Thus, we can rewrite \( I_2 \):
\[ I_2 = \int_{0}^{1} t e^{-t} \, dt \]

### Step 4: Combine the integrals
Now we have:
\[ I = I_1 + I_2 = \int_{-1}^{0} -t e^{-t} \, dt + \int_{0}^{1} t e^{-t} \, dt \]

### Step 5: Evaluate the integrals using integration by parts
Using integration by parts for both integrals:
1. For \( I_1 \):
Let \( u = -t \) and \( dv = e^{-t} dt \).
Then \( du = -dt \) and \( v = -e^{-t} \).
Thus,
\[ I_1 = \left[-t e^{-t}\right]_{-1}^{0} + \int_{-1}^{0} e^{-t} dt = [0 - (-1)e^{-(-1)}] + [1 - e^{-1}] \]
\[ = e^{-1} + 1 - e^{-1} = 1 \]

2. For \( I_2 \):
Similarly,
\[ I_2 = \left[t e^{-t}\right]_{0}^{1} - \int_{0}^{1} e^{-t} dt = [1 \cdot e^{-1} - 0] - [1 - e^{-1}] = e^{-1} - (1 - e^{-1}) = 2e^{-1} - 1 \]

### Step 6: Combine results
Now we can combine:
\[ I = 1 + (2e^{-1} - 1) = 2e^{-1} \]

### Final Result
Thus, the value of the integral is:
\[ I = 2(1 - \frac{1}{e}) \]