Question

The value of the integral \(\int _{1/e}^e\frac{|logx|}{x^2}dx\) is

• A. \(\frac{e^2-1}{2e}\)
• B. \(\frac{2}{e}\)
• C. \(2-\frac{2}{e}\)
• D. \(1-\frac{1}{e}\)

Hint:

Split the interval at x = 1 where log x changes sign, and integrate log x over x squared by parts.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\log x<0\) for \(x<1\) and \(\log x\ge0\) for \(x\ge1\). So we split the integral at \(x=1\).

Step 2: Antiderivative
By parts:
\[ \int\frac{\log x}{x^2}dx=-\frac{\log x}{x}-\frac1x=-\frac{\log x+1}{x} \]

Step 3: Part from 1/e to 1
Here \(|\log x|=-\log x\), so the integral is \(\left[\frac{\log x+1}{x}\right]_{1/e}^{1}=\frac{1}{1}-\frac{0}{1/e}=1\).

Step 4: Part from 1 to e
\[ \left[-\frac{\log x+1}{x}\right]_1^e=-\frac2e+1 \]

Step 5: Total
\[ 1+1-\frac2e=2-\frac2e \]
This is option (C).

Final Answer:
The two halves give 1 and 1 - 2/e, so the integral is 2 - 2/e, option (C). \[ \boxed{2-\frac{2}{e}} \]