I thought about this trick first and then found an example to apply it to.

$$\frac{dy}{dx}=1+\frac{2}{x+y}$$

$$\frac{dy}{dx}+\frac{dx}{dx}=1+\frac{2}{x+y}+1$$

$$\frac{d(x+y)}{dx}=2\frac{1+(x+y)}{x+y}$$

$$\frac{(x+y+1-1)d(x+y)}{1+(x+y)}=2dx$$

$$\frac{d(1+x+y)}{1+(x+y)}=d(y-x)$$

$$\ln|1+x+y|=(y-x)+\ln C$$

$$1+x+y=C\exp\left(y-x\right)$$

I am still not sure if this is a one-off case or if it can be generalised to a consistent method. It all spins around the understanding of \(d\) as a linear operator when it comes up in equations and integrals. This is something I discuss at greater length in my self-published book

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