Official solution
\[\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} \] + xy\[{{dy} \over {dx}}\] = 0
\[ \Rightarrow \] \[\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)} \] + xy\[{{dy} \over {dx}}\] = 0
\[ \Rightarrow \] \[\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)} \] = -xy\[{{dy} \over {dx}}\]
\[ \Rightarrow \] \[\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx\] .....(1)
Now put 1 + x2
= u2
and 1 + y2
= v2
2xdx = 2udu and 2ydy = 2vdv
\[ \Rightarrow \] xdx = udu and ydy = vdv
Substitute these values in equation (1)
\[\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}} \]
\[ \Rightarrow \] \[\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du\]
\[ \Rightarrow \] v = \[ - \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du\]
\[ \Rightarrow \] v = -u - \[{1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|\] + C
\[ \Rightarrow \] \[\sqrt {1 + {y^2}} \] = \[ - \sqrt {1 + {x^2}} \] + \[{1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|\] + C
\[ \Rightarrow \] \[\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C\]