The given differential equation is:

$x \frac{dy}{dx} + y \ln(x) = y \ln(y) + y$

Step 1: Simplify the equation

First, divide the entire equation by $y$ (assuming $y \neq 0$):

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

Step 2: Separate the variables

Rearrange the terms to separate the variables:

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

Recognize that $\ln(y) - \ln(x) = \ln\left(\frac{y}{x}\right)$, so the equation becomes:

$\frac{x}{y} \frac{dy}{dx} = \ln\left(\frac{y}{x}\right) + 1$

Step 3: Change of variables

Let's introduce a substitution to simplify the equation. Define:

$v = \frac{y}{x}$

Then, $y = vx$, and differentiating with respect to $x$:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substitute into the equation:

$\frac{x}{vx} \left(v + x \frac{dv}{dx}\right) = \ln(v) + 1$

Simplify:

$\frac{1}{v}(v + x \frac{dv}{dx}) = \ln(v) + 1$

$1 + \frac{x}{v} \frac{dv}{dx} = \ln(v) + 1$

Subtract 1 from both sides:

$\frac{x}{v} \frac{dv}{dx} = \ln(v)$

Step 4: Separate and integrate

Separate the variables:

$\frac{1}{\ln(v)} dv = \frac{1}{x} dx$

Now, integrate both sides:

$\int \frac{1}{\ln(v)} dv = \int \frac{1}{x} dx$

The integral on the right-hand side is straightforward:

$\int \frac{1}{x} dx = \ln|x| + C_1$

The integral on the left-hand side is a special function, known as the logarithmic integral, but in a simpler form it can be solved as:

$\text{Li}(v) = \ln|x| + C$

Final Answer:

The general solution of the differential equation is: In implicit form using special functions.

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