The Solution of differential equation

$\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$ is

Given that

$\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$

Here, $P = \frac{{2x}}{{1 + {x^2}}}$
and $Q = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$

which is a linear differential equation.
$\therefore$

${\rm{IF}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}} dx}}$

Put $1 + {x^2} = t \Rightarrow 2xdx = dt$

$\therefore {\rm{IF}} = {e^{\int {\frac{{dt}}{t}} }} = {e^{\log t}} = {e^{\log \left( {1 + {x^2}} \right)}} = 1 + {x^2}$

The general Solution is
$y \cdot \left( {1 + {x^2}} \right) = \int {\left( {1 + {x^2}} \right)} \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} + C$

$\Rightarrow$ $y\left( {1 + {x^2}} \right) = \int {\frac{1}{{1 + {x^2}}}} dx + C$

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