[tex]\frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\[/tex] [tex]t=\arctan y\\[/tex] [tex]t'=\frac{1}{1+y^2}y'\\[/tex] [tex]\frac{x}{\ln (x)t-1}=\frac{t}{t'}\\[/tex] [tex]\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\[/tex] [tex]xdt=(\ln (x)t-1)tdx\\[/tex] [tex]\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\[/tex] still cant beak it as one type of variable on each side so i substitute by another variable [tex]z=\ln x\\[/tex] [tex]dz=\frac{dx}{x}\\[/tex] [tex]\frac{dt}{zt-1}=dz\\[/tex] i dont know how to separate each variable type on one side so i could integrate ??
First of all, your last substitution should result in [tex]t'+t=zt^2[/tex]I'm not exactly sure on a good substitution to use, but since this is a Bernoulli ODE, you can use that method.