Question

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

• A. \[ \frac32 \]
• B. \[ \frac52 \]
• C. \[ 2 \]
• D. \[ 3 \]

Hint:

Whenever an integral contains \[ |\ln x|, \] split the interval at \[ x=1, \] because \(\ln x\) changes sign there.

Answer 1

The Correct Option is B

Concept: The sign of \[ \frac{\ln x}{x} \] depends only on \(\ln x\), since \(x>0\) on the given interval. Hence we split the integral at \[ \ln x=0 \quad\Longrightarrow\quad x=1. \]

Step 1: Remove the modulus. For \[ \frac1e\le x<1, \] \[ \ln x<0, \] so \[ \left|\frac{\ln x}{x}\right| = -\frac{\ln x}{x}. \] For \[ 1\le x\le e^2, \] \[ \ln x>0, \] so \[ \left|\frac{\ln x}{x}\right| = \frac{\ln x}{x}. \] Therefore, \[ I = -\int_{1/e}^{1}\frac{\ln x}{x}\,dx + \int_{1}^{e^2}\frac{\ln x}{x}\,dx. \]

Step 2: Use the standard integral. \[ \int \frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2}. \] Hence \[ I = -\left[ \frac{(\ln x)^2}{2} \right]_{1/e}^{1} + \left[ \frac{(\ln x)^2}{2} \right]_{1}^{e^2}. \]

Step 3: Evaluate each part. Since \[ \ln\!\left(\frac1e\right)=-1, \qquad \ln 1=0, \qquad \ln(e^2)=2, \] we get \[ I = -\left(0-\frac12\right) + \left(\frac{4}{2}-0\right). \] \[ = \frac12+2. \] \[ = \frac52. \]

Step 4: Write the final answer. \[ \boxed{\frac52} \]