$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$
$x(y + 1) \, dx = (x^2 + 1)(y^2 + 1) \, dy = 0$
$\dfrac{x(y + 1) \, dx}{(x^2 + 1)(y + 1)} = \dfrac{(x^2 + 1)(y^2 + 1) \, dy}{(x^2 + 1)(y + 1)} = 0$
$\dfrac{x \, dx}{x^2 + 1} = \dfrac{(y^2 + 1) \, dy}{y + 1} = 0$
By long division
$\dfrac{y^2 + 1}{y + 1} = y - 1 + \dfrac{2}{y + 1}$
Thus,
$\dfrac{x \, dx}{x^2 + 1} = \left( y - 1 + \dfrac{2}{y + 1} \right) \, dy = 0$
$\displaystyle \dfrac{1}{2} \int \dfrac{2x \, dx}{x^2 + 1} = \int y \, dy - \int dy + 2 \int \dfrac{dy}{y + 1} = 0$
$\frac{1}{2} \ln (x^2 + 1) = \frac{1}{2}y ^2 - y + 2 \ln (y + 1) + 2 \ln c = 0$
$\ln (x^2 + 1) = y ^2 - 2y + 4 \ln (y + 1) + 4 \ln c = 0$
$\ln (x^2 + 1) = y ^2 - 2y + 4 \, [ \, \ln (y + 1) + \ln c \, ] = 0$
$\ln (x^2 + 1) = y ^2 - 2y + 4 \ln | c(y + 1) | = 0$ answer
