Put $x^{2}=t \Rightarrow 2 x d x=d t$
$I=\int \frac{e^{x^{2}}\left(2+x^{2}\right) x d x}{\left(3+x^{2}\right)^{2}}=\frac{1}{2} \int e^{t} \frac{(2+t)}{(3+t)^{2}} d t$
$=\frac{1}{2} \int \frac{e^{t}(3+t-1)}{(3+t)^{2}} d t=\frac{1}{2} \int e^{t}\left[\frac{1}{3+t}-\frac{1}{(3+t)^{2}}\right] d t$
$=\frac{1}{2} e^{t} \frac{1}{3+t}+k\left[\because \frac{d}{d t}\left(\frac{1}{3+t}\right)=\frac{-1}{(3+t)^{2}}\right]$
$=\frac{1}{2} \frac{e^{x^{2}}}{3+x^{2}}+k$