Question:medium

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

Show Hint

Whenever an integral contains \[ |\ln x|, \] split the interval at \[ x=1, \] because \(\ln x\) changes sign there.
Updated On: Jul 9, 2026
  • \[ \frac32 \]
  • \[ \frac52 \]
  • \[ 2 \]
  • \[ 3 \]
Show Solution

The Correct Option is B

Solution and Explanation

Concept: Split integral at \(x=1\) where \(\ln x\) changes sign. Integrate \(\frac{\ln x}{x}\) using substitution.

Step 1:
For \(x\in[1/e,1)\), \(\ln x<0\): \(|\ln x|/x = -\ln x/x\). For \(x\in[1,e^2]\), \(\ln x\ge0\): \(|\ln x|/x = \ln x/x\). Integral = \(-\int_{1/e}^1 \frac{\ln x}{x}dx + \int_1^{e^2} \frac{\ln x}{x}dx\).

Step 2:
\(\int \frac{\ln x}{x}dx = \frac{(\ln x)^2}{2}\). Evaluate: \(-\left[0 - \frac{1}{2}\right] + \left[\frac{4}{2} - 0\right] = \frac12 + 2 = \frac52\).

Step 3:
Write the final answer. \(\boxed{\frac52}\)
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