Solution of the differential equation $\frac{dy}{dx}=\sqrt{1 + x^2 + y^2 + x^2y^2}$ is: (Here C is an arbitrary costant)

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

The correct answer is Option (3) → $\log\left|\frac{y+\sqrt{1+y^2}}{\sqrt{x+\sqrt{1+x^2}}}\right|=\frac{x}{2}\sqrt{1+x^2}+C$

Given differential equation:

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

Rewrite the expression under the square root:

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

So,

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

Rewrite as:

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

Integrate both sides:

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

The left integral is:

$\int \frac{dy}{\sqrt{1 + y^2}} = \sinh^{-1} y + C_1 = \ln \left( y + \sqrt{1 + y^2} \right) + C_1$

The right integral is:

$\int \sqrt{1 + x^2} \, dx = \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln \left| x + \sqrt{1 + x^2} \right| + C_2$

Combining constants $C = C_2 - C_1$, the implicit solution is:

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