The question is in the form of standard linear differential equation i.e. $\frac{dy}{dx}+yP\left ( x \right )=Q\left ( x \right )$,
for which the integrating factor is $I=e^{\int P\left ( x \right )dx}$.
For the given question, converting to standard form, $\frac{dy}{dx}+y\left ( -\frac{1}{x} \right )=1$
So, the Integrating factor would be: $e^{\int \left ( -\frac{1}{x} \right )dx}=\frac{1}{x}$.
$\therefore$ Solution to such a Differential equation is : $ye^{\int P\left ( x \right )dx}=\int \left \{ Q\left ( x \right )e^{\int P\left ( x \right )dx} \right \}dx$
Now, using the above solution definition, $y\left ( \frac{1}{x} \right )=\int \left ( 1 \right )\left ( \frac{1}{x} \right )dx$
$\Rightarrow$ $\frac{y}{x}=log_{e}x+c$ $\Rightarrow$ $y=x\left ( log_{e}x+c \right )$
Option A is the correct answer.