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

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$\begin{array}{ll} & \frac{\mathrm{d} y}{\mathrm{~d} x}+y=\frac{1+y}{x} \ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=\frac{1}{x}+\frac{y}{x} \ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}+\left(1-\frac{1}{x}ight) y=\frac{1}{x} \ 	herefore \quad & 	ext { I.F. }=\mathrm{e}^{\int\left(1-\frac{1}{x}ight) \mathrm{dx}}=\mathrm{e}^{x-\log x}=\frac{\mathrm{e}^x}{x}\end{array}$

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

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