The general solution of $\frac{d y}{d x}=1+x^2+y^2+x^2 y^2$ is: (given that C is the constant of integration)

$\tan ^{-1} y=x+\frac{x^3}{3}+C$

$\frac{d y}{d x}=\left(1+x^2\right)+y^2+x^2 y^2$

$\Rightarrow \frac{d y}{d x}=\left(1+x^2\right)+y^2\left(1+x^2\right) \Rightarrow \frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)$

$\Rightarrow \frac{1}{1+y^2} d y=\left(1+x^2\right) d x$

integrating both sides

we get $\int \frac{1}{1+y^2} d y=\int\left(1+x^2\right) d x$

$\tan ^{-1}(y)=\frac{x^3}{3}+x+C$