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: Approach
Use the substitution $x=e^{t}$ to remove the logarithm.

Step 2: Substitution
$dx=e^t\,dt$, and the limits go from $t=-1$ to $t=1$:
\[ \int_{-1}^{1}\frac{|t|}{e^{2t}}e^{t}dt=\int_{-1}^{1}|t|e^{-t}dt \]

Step 3: Split
$\int_0^1te^{-t}dt=1-\dfrac2e$. For $t\in[-1,0]$ use $|t|=-t$: $\int_{-1}^0(-t)e^{-t}dt$. Put $s=-t$: $\int_0^1se^{s}ds=1$.

Step 4: Total
$1+1-\dfrac2e=2-\dfrac2e$, option (C).

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