40. Solve the equation $x\left(y^{2}-1\right) d x+y\left(x^{2}+1\right) d y=0$.
Solution. Dividing all terms of the equation by the product $(x^{2}+1)\left(y^{2}-1\right)$, we get $\frac{x d x}{x^{2}+1}+\frac{y d y}{y^{2}-1}=0$. Now the variables are separated; integrating, we find

$$
\int \frac{x d x}{x^{2}+1}+\int \frac{y d y}{y^{2}-1}=C_{1} ; \frac{1}{2} \ln \left(x^{2}+1\right)+\frac{1}{2} \ln \left(y^{2}-1\right)=\frac{1}{2} \ln C
$$

Here the arbitrary constant $C_{1}$ is replaced by $\frac{1}{2} \ln C$ (since any positive or negative number can be represented as the natural logarithm of another, positive number $|C|$).

Dividing all terms of the equation by $1 / 2$, we get $\ln \left(x^{2}+1\right)\left(y^{2}-1\right)=$ $=\ln C$, from which $\left(x^{2}+1\right)\left(y^{2}-1\right)=C$. This is the general integral or the general solution of the given differential equation.

Remark. When dividing by the product $\varphi(x) F(y)$, we can lose those solutions of the equation that make this product zero (if they are part of the factor being canceled). $=x y^{2}$.