The general solution of the differential equation $x(1 + y^2)dx + y(1 + x^2)dy = 0$ is

The differential equation is $x(1 + y^2)dx + y(1 + x^2)dy = 0$

Rearranging to separate variables:

$x(1 + y^2)dx = -y(1 + x^2)dy$

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

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Integrating both sides:

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

For the left side, let $u = 1 + x^2$, then $du = 2x \, dx$:

$\int \frac{x}{1 + x^2}dx = \frac{1}{2}\ln|1 + x^2|$

For the right side, let $v = 1 + y^2$, then $dv = 2y \, dy$:

$-\int \frac{y}{1 + y^2}dy = -\frac{1}{2}\ln|1 + y^2|$

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Combining the results:

$\frac{1}{2}\ln|1 + x^2| = -\frac{1}{2}\ln|1 + y^2| + C_1$

$\ln|1 + x^2| = -\ln|1 + y^2| + C_2$

$\ln|1 + x^2| + \ln|1 + y^2| = C_2$

$\ln|(1 + x^2)(1 + y^2)| = C_2$

$(1 + x^2)(1 + y^2) = e^{C_2}$

Since $e^{C_2}$ is an arbitrary constant, let $C = e^{C_2}$:

$(1 + x^2)(1 + y^2) = C$