How to solve $y(\cos x+\ln y )+(x+ye^y)y'=0$?

Question

Find the general solution of the following differential equation: $$y(\cos(x)+\ln(y))+(x+ye^y)y'=0$$

I have not even been able to change this equation to a better form.

Answer 1

$$y(\cos(x)+\ln(y))+(x+ye^y)y'=0\\\cos(x)+\ln(y)+\frac{xy'}y+y'e^y=0\\\left(\sin x\right)'+\ln y+x(\ln y)'+\left(e^y\right)'=0\\\left(\sin x+e^y\right)'+(x\ln y)'=0\\\sin x+e^y+x\ln y=c$$where $c$ is a constant. The trick here is to try and rearrange things to give terms which look like the results of chain rule or product rule.