${{x}^{2}}+3x+1=(x+3)\sqrt{{{x}^{2}}+1}$
$\begin{align}
& {{x}^{2}}+3x+1-(x+3)\sqrt{{{x}^{2}}+1}=0 \\
& ({{x}^{2}}+1)+(3x+9)-9-(x+3)\sqrt{{{x}^{2}}+1}=0 \\
& ({{x}^{2}}+1)+3(x+3)-9-(x+3)\sqrt{{{x}^{2}}+1}=0 \\
& \left[ \left( {{x}^{2}}+1 \right)-9 \right]-\left[ (x+3)\sqrt{{{x}^{2}}+1}-3(x+3) \right]=0 \\
& \left( \sqrt{{{x}^{2}}+1}-3 \right)\left( \sqrt{{{x}^{2}}+1}+3 \right)-\left( x+3 \right)\left( \sqrt{{{x}^{2}}+1}-3 \right)=0 \\
& \left( \sqrt{{{x}^{2}}+1}-3 \right)\left[ \left( \sqrt{{{x}^{2}}+1}+3 \right)-\left( x+3 \right) \right]=0 \\
& \left( \sqrt{{{x}^{2}}+1}-3 \right)\left( \sqrt{{{x}^{2}}+1}+3-x-3 \right)=0 \\
& \left( \sqrt{{{x}^{2}}+1}-3 \right)\left( \sqrt{{{x}^{2}}+1}-x \right)=0 \\
\end{align}$
$\left[ \begin{align}
& \sqrt{{{x}^{2}}+1}-3=0 \\
& \sqrt{{{x}^{2}}+1}-x=0 \\
\end{align} \right.$ $\Rightarrow $ $\left[ \begin{align}
& \sqrt{{{x}^{2}}+1}=3 \\
& \sqrt{{{x}^{2}}+1}=x(x>0) \\
\end{align} \right.$
$\Rightarrow $$\left[ \begin{align}
& {{x}^{2}}+1=9\Rightarrow {{x}^{2}}=8\Rightarrow x=\pm 2\sqrt{2} \\
& {{x}^{2}}+1={{x}^{2}}(loai) \\
\end{align} \right.$
Vậy phương trình có hai nghiệm : $x=2\sqrt{2}$ và $x=-2\sqrt{2}$