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$\Large \int_{e^-1}^{e^2} \Biggm| \dfrac{log_e (x)}{x}  \Biggm| dx$

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 $\Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$

∵ $\Large\int_a^b f(x) dx = F(b)-F(a)$

$\qquad\qquad =\displaystyle\lim_{x\to b-} (F(x))- \displaystyle\lim_{x\to a+} (F(x))$

$\displaystyle\lim_{x\to e^2-}(\dfrac{1}{2} \ln^2(x))$

$= \displaystyle\lim_{x\to 1-}-\dfrac{1}{2} \ln^2(x)$

$= \dfrac{1}{2}ln^2 (e^2)$

$= \dfrac{1}{2}.2^2$ $\qquad\qquad\Bigg[∵ln(e^2) = 2.ln(e) = 2 \\\qquad\qquad∴ln^2(e^2) = 2^2 =4\Bigg]$

$= 2$

$\displaystyle\lim_{x\to 1+}(\dfrac{1}{2} \ln^2(x))$

$= \dfrac{1}{2}. ln^2 (1)$ $\qquad \Bigg[∵ln(1) =0\\ \qquad ln^2(1) = 0\Bigg]$

$=\dfrac{1}{2} . 0$

$=0$

∴$\displaystyle\lim_{x\to e^2-}(\dfrac{1}{2} \ln^2(x)) - \displaystyle\lim_{x\to 1+}(\dfrac{1}{2} \ln^2(x))$

$= 2-0$

$=2$

$∴ \Large \int _{e^{-1}}^1 ( -\dfrac{ln(x)}{x}dx ) + \Large \int _1^{e^2} ( \dfrac{log_e(x)}{x}dx )$

$= \dfrac{1}{2} + 2$

$=\color{orange}{ 2.5}$
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$\begin{vmatrix} log_{e}x/x \end{vmatrix} = \begin{Bmatrix} - log_{e}x/x & 0<x < 1\\ \\ log_{e}x/x & x \geqslant 1 \end{Bmatrix}$

 

$-\int_{e^{-1}}^{1}log_{e}x/x .dx\ + \int_{1}^{e^2}log_{e}x/x.dx$

let,     

$log_{e} x = t$

$1/x. dx = dt$

$=-\int_{-1}^{0}tdt \ + \int_{0}^{2}tdt$

$=-{t^2}/2\ \left.\begin{matrix} 0\\-1 \end{matrix}\right| \ + \ {t^2}/2\ \left.\begin{matrix} 2\\0 \end{matrix}\right|$

$= -(0 - (-1)^2/2 ) + ( 0 + (2^2)/2)$

$= -(0 - 1/2) + ( 0 + 2)$

$= 1/2 + 2$

$=2.5$
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