Solve the differential equation $\left( {1 + {y^2}} \right){\tan ^{ - 1}}xdx + 2y\left( {1 + {x^2}} \right)dy = 0$.

Given differential equation is
$\left( {1 + {y^2}} \right){\tan ^{ - 1}}xdx + 2y\left( {1 + {x^2}} \right)dy = 0$

$\Rightarrow$ $\left( {1 + {y^2}} \right){\tan ^{ - 1}}xdx = - 2y\left( {1 + {x^2}} \right)dy$

$\Rightarrow$ $\frac{{{{\tan }^{ - 1}}xdx}}{{1 + {x^2}}} = - \frac{{2y}}{{1 + {y^2}}}dy$

On integrating both sides,

we get
$\int {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}} dx = - \int {\frac{{2y}}{{1 + {y^2}}}} dy$

Put ${\tan ^{ - 1}}x = t$ in LHS,

we get $\frac{1}{{1 + {x^2}}}dx = dt$

and put $1 + {y^2} = u$ in RHS,

we get $2ydy = du$

$\Rightarrow$ $\int t dt = - \int {\frac{1}{u}} du \Rightarrow \frac{{{t^2}}}{2} = - \log u + C$

$\Rightarrow$ $\frac{1}{2}{\left( {{{\tan }^{ - 1}}x} \right)^2} = - \log \left( {1 + {y^2}} \right) + C$

$\Rightarrow$ $\frac{1}{2}{\left( {{{\tan }^{ - 1}}x} \right)^2} + \log \left( {1 + {y^2}} \right) = C$