Example 5. Solve the equation: $2 x+1+x \sqrt{x^{2}+2}$ $+(x+1) \sqrt{x^{2}+2 x+3}=0$.
Thus, we have
\[
\left\{\begin{array}{l}
2 x+1+u x+(x+1) v=0, \\
v^{2}-u^{2}=2 x+1 .
\end{array}\right.
\]
Solving this system of equations for \( u \) and \( v \) is simpler than directly solving the original equation, but it is still a quadratic equation, making it somewhat complicated. Therefore, consider further expressing the parameter \( x \) in terms of \( u \) and \( v \), so that the solutions for \( u \) and \( v \) are constants.
From (2), \( x=\frac{v^{2}-u^{2}-1}{2} \).
Thus, \( x+1=\frac{v^{2}-u^{2}+1}{2} \).
Substituting (2) and (3) into (1), we get
\[
v^{2}-u^{2}+\frac{v^{2}-u^{2}-1}{2} \cdot u+\frac{v^{2}-u^{2}+1}{2} \cdot v=0.
\]

Rearranging and factoring out common terms, we get
\[
\begin{array}{l}
(v-u)\left[2(v+u)+(u+v)^{2}+1\right]=0. \\
\because u, v>0, \\
\therefore 2(v+u)+(v+u)^{2}+1 \neq 0. \\
\therefore v=u,
\end{array}
\]

Thus, \( \sqrt{x^{2}+2}=\sqrt{x^{2}+2 x+3} \).
Solving for \( x \), we get \( x=-\frac{1}{2} \).