Given, differential equation is
$\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}$
It is a linear equation of the form
$\frac{d y}{d x}+P y=Q$
where $P=\frac{1}{x \log _{\theta} x} $ and $ Q=\frac{1}{x}$
$\therefore $ Integrating factor, IF $=e$
$=e^{\int \frac{1}{x \log _{e} x} d x}$
$=e^{\log \left(\log _{e} x\right)}$
$=\log _{e} x$
$\therefore $ Solution of differential equation is
$y \times \log _{e} x=\int \frac{1}{x} \log _{e} x \,d x $
$\Rightarrow y \times \log _{e} x=\frac{\left(\log _{\theta} x\right)^{2}}{2}+C\,\,\,...(i)$
When $y=1$ and $x=e$, then
$1 \times \log _{e} e=\frac{\left(\log _{\theta} e\right)^{2}}{2}+C$
$\Rightarrow 1=\frac{1}{2}+C$
$\Rightarrow C=\frac{1}{2}$
On putting $C=\frac{1}{2}$ in Eq. (i), we get
$Y \times \log _{\theta} x=\frac{\left(\log _{\theta} x\right)^{2}}{2}+\frac{1}{2}$
$\Rightarrow 2 y=\log _{e} x+\frac{1}{\log _{e} x}$