$(1 + {x^2})dy + 2xydx = \cot xdx(x \ne 0)$

: The given equation is $(1 + {x^2})dy + 2xydx = \cot xdx$

$\Rightarrow \cfrac{{dy}}{{dx}} + \cfrac{2}{{1 + {x^2}}}y = \cfrac{{\cot x}}{{1 + {x^2}}}$

(1)
which is a linear equation of the type
where, $P = \cfrac{2}{{1 + {x^2}}}$ and $Q = \cfrac{{\cot x}}{{1 + {x^2}}}$

Now, $\int {Pdx} = \int {\cfrac{{2x}}{{1 + {x^2}}}dx} = \log \left| {1 + {x^2}} \right| = \log \left( {1 + {x^2}} \right)$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int {Pdx} }} = {e^{\log (1 + {x^2})}} = 1 + {x^2}$ $\therefore$
The solution is $y \cdot ({\rm{I}}{\rm{.F}}{\rm{.}}) = \int {Q \cdot ({\rm{I}}{\rm{.F}}{\rm{.}})dx} + C$

$\Rightarrow y(1 + {x^2}) = \int {\cot xdx} + C \Rightarrow y(1 + {x^2}) = \log \left| {\sin x} \right| + C$

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

which is the required solution.