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

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

The correct answer is Option (2) → $\frac{e^x}{x}$ ##

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

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

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

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

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

Which is a linear differential equation.

Now, compare with $\frac{dy}{dx} + P \cdot y = Q$

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

$\text{I.F} = e^{\int P \, dx} = e^{-\int \frac{1-x}{x} \, dx} = e^{\int \frac{x-1}{x} \, dx}$

$= e^{\int (1 - \frac{1}{x}) \, dx} = e^{x - \log x}$

$= e^x \cdot e^{\log(\frac{1}{x})} = e^x \cdot \frac{1}{x} \quad [∵e^{\log k} = k]$