Solution of the differential equation

$\frac{dy}{dx} = \sqrt{1 + x^2 + y^2 + x^2y^2}$ is : (Here $C$ is an arbitrary constant)

Given:

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

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The expression under the square root can be factored:

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

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

The equation becomes:

$\frac{dy}{dx} = \sqrt{(1+x^2)(1+y^2)}$

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

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Separating variables:

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

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Integrating both sides:

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

$\log|y + \sqrt{1+y^2}| = \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x + \sqrt{1+x^2}| + C$

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Rearranging:

$\log|y + \sqrt{1+y^2}| - \frac{1}{2}\log|x + \sqrt{1+x^2}| = \frac{x}{2}\sqrt{1+x^2} + C$

Using logarithm properties $\log a - \log b = \log\frac{a}{b}$ and $\frac{1}{2}\log a = \log\sqrt{a}$:

$\log\left|\frac{y + \sqrt{1+y^2}}{\sqrt{x + \sqrt{1+x^2}}}\right| = \frac{x}{2}\sqrt{1+x^2} + C$

Therefore:

$\log\left|\frac{y+\sqrt{1+y^2}}{\sqrt{x+\sqrt{1+x^2}}}\right| = \frac{x}{2}\sqrt{1+x^2} + C$