The general solution of the differential equation $$\sqrt{1 + x^2 + y^2 + x^2y^2} + xy\frac{dy}{dx} = 0$$ (where C is a constant of integration)

We start from the given differential equation

$$\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}\;+\;xy\,\frac{dy}{dx}=0.$$

The first task is to simplify the square-root term. Observe that

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

because expanding the right‐hand product gives $$1+y^{2}+x^{2}+x^{2}y^{2},$$ exactly the required expression. Hence

$$\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}} =\sqrt{(1+x^{2})(1+y^{2})} =\sqrt{1+x^{2}}\;\sqrt{1+y^{2}}.$$

Substituting this back, the differential equation becomes

$$\sqrt{1+x^{2}}\;\sqrt{1+y^{2}}\;+\;xy\,\frac{dy}{dx}=0.$$

We now isolate the derivative term:

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

Dividing by $$x\,y$$ (and assuming $$x\neq0,\;y\neq0$$ for separation) gives

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

To separate the variables we multiply both sides by $$\dfrac{y}{\sqrt{1+y^{2}}}$$ and by $$dx$$:

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

Now both sides involve a single variable, so we integrate:

Left integral:
We have to evaluate $$\displaystyle\int \frac{y}{\sqrt{1+y^{2}}}\,dy.$$ Put $$1+y^{2}=t\;\Longrightarrow\;dt=2y\,dy\;\Rightarrow\;y\,dy=\dfrac{dt}{2}.$$ Therefore

$$\int \frac{y}{\sqrt{1+y^{2}}}\,dy=\int\frac{1}{2}\,\frac{dt}{\sqrt{t}} =\frac{1}{2}\cdot2\,\sqrt{t} =\sqrt{t} =\sqrt{1+y^{2}}.$$

Right integral:
We need $$\displaystyle\int-\frac{\sqrt{1+x^{2}}}{x}\,dx.$$ First, set $$x=\sinh u.$$ Then $$dx=\cosh u\,du$$ and $$\sqrt{1+x^{2}}=\sqrt{1+\sinh^{2}u}=\cosh u.$$ Hence

$$-\int\frac{\sqrt{1+x^{2}}}{x}\,dx =-\int\frac{\cosh u}{\sinh u}\;\cosh u\,du =-\int\frac{\cosh^{2}u}{\sinh u}\,du.$$

Using the identity $$\cosh^{2}u=1+\sinh^{2}u,$$ we write

$$\frac{\cosh^{2}u}{\sinh u}=\frac{1}{\sinh u}+\sinh u.$$

Thus the right integral becomes

$$-\int\!\left(\frac{1}{\sinh u}+\sinh u\right)\!du =-\left(\int\operatorname{csch}u\,du+\int\sinh u\,du\right).$$

We use two standard results: (1) $$\displaystyle\int\operatorname{csch}u\,du=\ln\!\left|\tanh\frac{u}{2}\right|,$$ (2) $$\displaystyle\int\sinh u\,du=\cosh u.$$ Therefore

$$-\int\frac{\sqrt{1+x^{2}}}{x}\,dx =-\left(\ln\!\left|\tanh\frac{u}{2}\right|+\cosh u\right)+C_1,$$

where $$C_1$$ is a constant of integration. Returning to $$x$$ (with $$x=\sinh u$$ and $$\cosh u=\sqrt{1+x^{2}}$$) we have

$$-\int\frac{\sqrt{1+x^{2}}}{x}\,dx =-\sqrt{1+x^{2}}-\ln\!\left|\tanh\frac{u}{2}\right|+C_1.$$ But $$\displaystyle\tanh\frac{u}{2}=\frac{\sinh u}{\cosh u+1} =\frac{x}{\sqrt{1+x^{2}}+1}.$$ Hence

$$-\int\frac{\sqrt{1+x^{2}}}{x}\,dx =-\sqrt{1+x^{2}}-\ln\!\left|\frac{x}{\sqrt{1+x^{2}}+1}\right|+C_1.$$

Collecting the two integrated parts, we obtain

$$\sqrt{1+y^{2}} =-\sqrt{1+x^{2}} -\ln\!\left|\frac{x}{\sqrt{1+x^{2}}+1}\right| +C_1.$$

Move the two negative terms to the left side to tidy the expression:

$$\sqrt{1+y^{2}}+\sqrt{1+x^{2}} =-\ln\!\left|\frac{x}{\sqrt{1+x^{2}}+1}\right|+C_1.$$

Because a logarithm with a minus sign is the logarithm of the reciprocal, we write

$$-\ln\!\left|\frac{x}{\sqrt{1+x^{2}}+1}\right| =\ln\!\left|\frac{\sqrt{1+x^{2}}+1}{x}\right|.$$

Next, we use the property $$\ln a = \tfrac12 \ln a^{2}$$ to switch to a more symmetrical form. Squaring the argument inside the log gives

$$\left(\frac{\sqrt{1+x^{2}}+1}{x}\right)^{2} =\frac{(\sqrt{1+x^{2}}+1)^{2}}{x^{2}} =\frac{(\sqrt{1+x^{2}}+1)}{(\sqrt{1+x^{2}}-1)},$$ because $$x^{2}=(\sqrt{1+x^{2}})^{2}-1$$ and cancellation leaves the stated ratio. Therefore

$$\ln\!\left|\frac{\sqrt{1+x^{2}}+1}{x}\right| =\frac12\ln\!\left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right|.$$

Replacing the logarithm in our equation with this equivalent gives the compact final relation

$$\sqrt{1+y^{2}}+\sqrt{1+x^{2}} =\frac12\,\log_e\!\left(\frac{\sqrt{1+x^{2}}+1} {\sqrt{1+x^{2}}-1}\right)+C,$$ where $$C$$ absorbs the earlier constant $$C_1.$$

This matches exactly the expression offered in Option C.

Hence, the correct answer is Option C.