Integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$, is

$\frac{x}{e^x}$ $\frac{e^x}{x}$ $x e^x$ $e^x$

$\frac{e^x}{x}$

We have, $\frac{d y}{d x}+\left(1-\frac{1}{x}\right) y=\frac{1}{x}$ Integrating factor = $e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x-\log x}=\frac{1}{x} e^x$