The solution of the differential equation $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{\left(1+x^2\right)}^2}$ is 

We have, $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{\left(1+x^2\right)}^2}$

Clearly, it is a linear differential equation of the form

$\frac{dy}{dx}+Py=Q$ where $P=\frac{2x}{1+x^2}$ and $Q=\frac{1}{{\left(1+x^2\right)}^2}$

$\text{ I.F. }=e^{\int Pdx}=e^{\log \left(1+x^2\right)}=\left(1+x^2\right)$

Multiplying both sides of (i) by $\left(1+x^2\right)$ and integrating, we get 

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