$\Large \int_{e^-1}^{e^2} \Biggm| \dfrac{log_e (x)}{x} \Biggm| dx$
In first step, we'll try to eliminate the absolutes
We'll try to find
the equivalent expressions to $ \Biggm| \dfrac{log_e (x)}{x} \Biggm|$ at $e^{-1}\leq x \leq e^2$ without the absolutes
at $e^{-1}\leq x \leq 1$ $:$ $-\dfrac{ln(x)}{x}$
at $1\leq x \leq e^2$ $:$ $-\dfrac{log_e(x)}{x}$
This can be written as
$= \Large \int _{e^{-1}}^1 ( -\dfrac{ln(x)}{x}dx ) + \Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$
$\Large\int _{e^{-1}}^1 -\dfrac{ln(x)}{x}dx$
First calculate $\Large\int -\dfrac{ln(x)}{x}dx$
We know that $\Large\int a.f(x) dx = a\Large\int f(x) dx$
∴ $\Large\int -\dfrac{ln(x)}{x}dx = -\Large\int \dfrac{ln(x)}{x}dx$
Now, taking $ln(x) = u$
∴ $\dfrac{d}{dx}(ln(x) )= \dfrac{du}{dx}$
Or, $\dfrac{1}{x} =\dfrac{du}{dx}$
Or, $dx = x du$
∴ $-\Large\int \dfrac{ln(x)}{x}dx $
= $-\Large\int u. du$
= $-\Large\int \dfrac{u^{1+1}}{1+1}du$
= $\dfrac{1}{2}u^2+c$ $\qquad \text{By applying power rule} \Large\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$
Now, putting back $u = ln(x)$
∴ $-\dfrac{1}{2}u^2 + c =-\dfrac{1}{2} \ln^2(x) + c $
Now, $\Large \int _{e^{-1}}^1 ( -\dfrac{ln(x)}{x}dx )$
We know that, $\Large\int_a^b f(x) dx = F(b)-F(a)$
$\qquad\qquad =\displaystyle\lim_{x\to b-} (F(x))- \displaystyle\lim_{x\to a+} (F(x))$
Now, compute $\displaystyle\lim_{x\to b-} (F(x))$
$\qquad = \displaystyle\lim_{x\to 1-}-\dfrac{1}{2} \ln^2(x) $
$\qquad = -\dfrac{1}{2} \ln^2(1) $
$\qquad = -\dfrac{1}{2} .0 $ $\qquad \Big[As\hspace{0.1cm} ln(1) =0\\ \qquad∴ ln^2 (1) = 0\Big]$
$\qquad = 0$
Compute $\displaystyle\lim_{x\to a+} (F(x))$
$= \displaystyle\lim_{x\to e^{-1}+} -\dfrac{1}{2} \ln^2(x)$
$= -\dfrac{1}{2} \ln^2(e^{-1})$
$= -\dfrac{1}{2} .1$
$∵\bigg[We\hspace{0.1cm} know, ln(e^{-1}) = -1.ln(e) [∵ log_a(x^b) = b.(log_a(x))]\hspace{0.1cm}\bigg]$
$= -\dfrac{1}{2} $
Now, $\Large \int _{e^{-1}}^1 ( -\dfrac{ln(x)}{x}dx ) $
$= \displaystyle\lim_{x\to b-} (-\dfrac{1}{2} \ln^2(x))- \displaystyle\lim_{x\to a+} (-\dfrac{1}{2} \ln^2(x))$
$= 0-(-\dfrac{1}{2})$
$= \dfrac{1}{2}$
Compute, $\Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$
Compute first, $\large\int( \dfrac{log_e(x)}{x}dx )$
We know that, $log_e(x) = ln(x)$
∴$=\large\int \dfrac{ln(x)}{x}dx $
Assuming,$ln(x) = u$
Or, $dx =xdu$
∴$\large\int \dfrac{ln(x)}{x}dx = \large\int udu$
$ \large\int udu = \dfrac{u^2}{2}+c$
putting back, $u = ln(x)$
∴$\large\int udu = \dfrac{ln^2(x)}{2}+c$
Now, we have to compute $\Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$
∵ $\Large\int_a^b f(x) dx = F(b)-F(a)$
$\qquad\qquad =\displaystyle\lim_{x\to b-} (F(x))- \displaystyle\lim_{x\to a+} (F(x))$
$\displaystyle\lim_{x\to e^2-}(\dfrac{1}{2} \ln^2(x))$
$= \displaystyle\lim_{x\to 1-}-\dfrac{1}{2} \ln^2(x)$
$= \dfrac{1}{2}ln^2 (e^2)$
$= \dfrac{1}{2}.2^2$ $\qquad\qquad\Bigg[∵ln(e^2) = 2.ln(e) = 2 \\\qquad\qquad∴ln^2(e^2) = 2^2 =4\Bigg]$
$= 2$
$\displaystyle\lim_{x\to 1+}(\dfrac{1}{2} \ln^2(x))$
$= \dfrac{1}{2}. ln^2 (1)$ $\qquad \Bigg[∵ln(1) =0\\ \qquad ln^2(1) = 0\Bigg]$
$=\dfrac{1}{2} . 0$
$=0$
∴$\displaystyle\lim_{x\to e^2-}(\dfrac{1}{2} \ln^2(x)) - \displaystyle\lim_{x\to 1+}(\dfrac{1}{2} \ln^2(x))$
$= 2-0$
$=2$
$∴ \Large \int _{e^{-1}}^1 ( -\dfrac{ln(x)}{x}dx ) + \Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$
$= \dfrac{1}{2} + 2$
$=\color{orange}{ 2.5}$