$\frac{dy}{dx}=1+\frac{2}{x+y}$ solution by an "integrating term" I though about this trick and then found an example to apply it to:
$$\frac{dy}{dx}=1+\frac{2}{x+y}$$
This is the trick: add  $\frac{dx}{dx}=1$ to both parts
$$\frac{dy}{dx}+\frac{dx}{dx}=1+\frac{2}{x+y}+1$$
Using the linearity of $d$
$$\frac{d(x+y)}{dx}=2\frac{1+(x+y)}{x+y}$$
$$\frac{(x+y)d(x+y)}{1+(x+y)}=2dx$$
$$d(x+y)-\frac{d(x+y)}{1+(x+y)}=2dx$$
$$-\frac{d(x+y)}{1+(x+y)}=2dx-d(x+y)$$
Now $2dx-d(x+y)=2dx-dx-dy=dx-dy=d(x-y)$
$$-\frac{d(x+y)}{1+(x+y)}=d(x-y)$$
$$\frac{d(1+x+y)}{1+(x+y)}=d(y-x)$$
Integrating:
$$\ln|1+x+y|=(y-x)+\ln C$$
$$1+x+y=C\exp\left(y-x\right)$$
Is this a one-off case, or a particular example of a certain method? Does anyone know more examples of ODE's that can be solved similarly? I know the integrating multiplier theory quite well, but this one seems like something extra to that.

In general for equations of the form $y'(x)=f(ax+by+c)$, (where $a, b$ and $c$ are constantes $b\neq 0$), you can use the change of variables $u=ax+by+c$. You obtain the equation:
$$
y'=\frac{u'-a}{b}=f(u)
$$
which can be solved using separation of variables:
$$
\frac{du}{b \,f(u)+a}=dx
$$
