How to solve $\frac{\mathrm dy}{\mathrm dx}=\frac{x+y}{x}$ I am looking for a help with the below equation
$$\frac{\mathrm dy}{\mathrm dx}=\frac{x+y}{x}$$
I don't get the suggested answer $$y = x\ln(x) + c x$$
My attempt
$$\mathrm dx(1+\frac{y}{x})=\mathrm dy$$
$$c+x+y\ln(x)=y$$
$$y=\frac{c+x}{1-\ln x}$$

The problem with your approach is that:
$$ \int \left(1 + \frac{y}{x}\right) dx \neq C + x + y \ln x $$
Note that $y$ is a function of $x$, so you cannot treat it as a constant when you integrate.
One appropriate way to do this problem using an integrating factor:
Doing some algebra:
$$ y'(x) - y(x)/x = 1 $$
Let $\mu(x) = \exp(\int -\frac1x dx) = 1/x$. Multiplying both sides by $\mu$:
$$ y'(x)/x - y(x)/x^2 = 1/x $$
Note that $-1/x^2$ is the derivative of $1/x$:
$$ \frac{dy}{dx} \frac{1}{x} + \frac{d}{dx} \left( \frac{1}{x} \right) y(x) = \frac{1}{x} $$
Using the product rule:
$$ \frac{d}{dx} (y/x) = \frac{1}{x}$$
You can finish the rest yourself.
