Answer
$\frac{1}{6}$
Work Step by Step
At first we need to calculate the intersection point.
The area is indicated by the yellow line.
Given that,
$y=\frac{ln x}{x}$ $,$ $y=\frac{(ln x)^2}{x}$
So,
$ \frac{ln x}{x} = \frac{(ln x)^{2}}{x} $
$ln x = (ln x)^2$
$ 0 = (ln x)^2 - ln x $
$ 0 = lnx(lnx -1) $
$ lnx =$ $0$ or $1 $
$x=$ $e^0$ $ $ or $e^1$ $ $ $[1 $ or $ e]$
Now,
$A = \int_{1}^{e}[\frac{ln x}{x} - \frac{(lnx)^2}{x}]dx $ $ = [\frac{1}{2}(lnx)^2 - \frac{1}{3}(ln x)^3]_{1}^{e}$
$= (\frac{1}{2}-\frac{1}{3})-(0-0)$
$=\frac{1}{6}$
