Ans:(A)

Explanation:

$$\begin{gathered} \mathrm{Given} \mathrm{that}, \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2\mathrm{xy}}{1+{\mathrm{x}}^2}=\frac{1}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \mathrm{Here}, \mathrm{P}=\frac{2\mathrm{x}}{1+{\mathrm{x}}^2} \mathrm{and} \mathrm{Q}=\frac{1}{{\left[1+{\mathrm{x}}^2\right]}^2} \\ \mathrm{Which} \mathrm{is} \mathrm{a} \mathrm{linear} \mathrm{differential} \mathrm{equation} \\ \therefore \mathrm{IF}={\mathrm{e}}^{\int \frac{\mathrm{dt}}{\mathrm{t}}}={\mathrm{e}}^{\log \mathrm{t}}={\mathrm{e}}^{\log \left(1+{\mathrm{x}}^2\right)}=1+{\mathrm{x}}^2 \\ \mathrm{The} \mathrm{general} \mathrm{solution} \mathrm{is} \\ \mathrm{y}.\left(1+{\mathrm{x}}^2\right)=\int \left(1+{\mathrm{x}}^2\right)\frac{1}{{\left(1+{\mathrm{x}}^2\right)}^2}\mathrm{dx}+\mathrm{c} \\ \Rightarrow \mathrm{y}\left(1+{\mathrm{x}}^2\right)=\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{c} \\ \Rightarrow \mathrm{y}\left(1+{\mathrm{x}}^2\right)={\tan }^{-1}\left(\mathrm{x}\right)+\mathrm{c} \end{gathered}$$