The integrating factor of differential equation $\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$ is

Given that, $\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{1 + y}}{x} - y$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{1 + y - xy}}{x}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{y(1 - x)}}{x}$

$\Rightarrow$ $\frac{{dy}}{{dx}} - \left( {\frac{{1 - x}}{x}} \right)y = \frac{1}{x}$

Here, $P = \frac{{ - (1 - x)}}{x},Q = \frac{1}{x}$

${\rm{IF}} = {e^{\int P dx}} = {e^{ - \int {\frac{{1 - x}}{x}} dx}} = {e^{\int {\frac{{x - 1}}{x}} dx}}$

$= {e^{\int {\left( {1 - \frac{1}{x}} \right)} dx}}$

$= {e^{\int x - \log x}}$

$= {e^x} \cdot {e^{\log \left( {\frac{1}{x}} \right)}}$

$= {e^x} \cdot \frac{1}{x}$