$\displaystyle \frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\$

$\displaystyle t=\arctan y\\$

$\displaystyle t'=\frac{1}{1+y^2}y'\\$

$\displaystyle \frac{x}{\ln (x)t-1}=\frac{t}{t'}\\$

$\displaystyle \frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\$

$\displaystyle xdt=(\ln (x)t-1)tdx\\$

$\displaystyle \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$

still cant beak it as one type of variable on each side

so i substitute by another variable

$\displaystyle z=\ln x\\$

$\displaystyle dz=\frac{dx}{x}\\$

$\displaystyle \frac{dt}{zt-1}=dz\\$

i dont know how to separate each variable type on one side

so i could integrate

??