The integrating factor of the differential equation dy/dx + y = 1+y/x is

The correct option is e^x/x.

To find the integrating factor of the given differential equation, we first need to rearrange it into the standard form of a first-order linear differential equation.

The given differential equation is:

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

We can rearrange the terms by bringing all the terms involving y to the left-hand side of the equation:
$$\frac{dy}{dx}+y-\frac{y}{x}=\mathrm{1<!--\; -->}$$

Factoring out y on the left-hand side gives us:
$$\frac{dy}{dx}+(1-\frac{1}{x})y=1$$

This is now in the standard first-order linear differential equation form:
$$\frac{dy}{dx}+P(x)y=Q(x)$$

Comparing the two equations, we identify the coefficient function P(x) as:

The integrating factor (I.F.) is calculated using the formula:
$$\text{I.F.}=e^{\int P(x)dx}$$

First, we evaluate the integral of P(x) with respect to x:
$$\int P(x)dx=\int (1-\frac{\mathrm{1<!--\; <mi-->x}}{})dx=x-\ln (x)$$

Now, we substitute this result back into the integrating factor expression:
$$\text{I.F.}=e^{x-\ln (x)}$$

Using the laws of exponents, we can split the expression in the exponent:
$$\text{I.F.}=e^x\cdot e^{-\ln (x)}$$

We know that:
$$e^{-\ln (x)}=e^{\ln (x^{-1})}=x^{-1}=\frac{\mathrm{1<!--\; <mi-->x}}{}$$

Substituting this back into the expression for the integrating factor gives:
$$\text{I.F.}=e^x\cdot \frac{\mathrm{1<!--\; <mi-->x}}{}=\frac{e^x}{x}$$

Thus, the integrating factor is e^x/x.