If I = _ 1 / e ^ e | x| d x x^2 , then I is equal to - Vidyadip

MCQ

If $I =\int_{1 / e}^{ e }|\log x| \frac{ d x}{x^2}$, then I is equal to

A
2
B
$\frac{2}{ e }$
✓
$2\left(1-\frac{1}{ e }\right)$
D
$0$

✓

Answer

Correct option: C.

$2\left(1-\frac{1}{ e }\right)$

(C)
Since $|\log x|=-\log x$, if $\frac{1}{ e }<x<1$
$=\log x$, if $1<x< e$
$\therefore I =\int_{1 / e }^1 \frac{(-\log x)}{x^2} d x+\int_1^{ e } \frac{\log x}{x^2} d x$
$=-\left[-\frac{\log x}{x}-\frac{1}{x}\right]_{1 / e }^1+\left[-\frac{\log x}{x}-\frac{1}{x}\right]_1^{ e } \\ =\left[0+1-\left(\frac{\log \frac{1}{ e }}{\frac{1}{ e }}+\frac{1}{\frac{1}{ e }}\right)\right]-\left[\frac{\log e }{ e }+\frac{1}{ e }-(0+1)\right] \\ =1-(- e + e )-\left(\frac{2}{ e }-1\right) \\ =2-\frac{2}{ e }=2\left(1-\frac{1}{ e }\right)$

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