2. Let real numbers $x, y$ satisfy
$$
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1 \text {. }
$$

Find the value of $x+y$.
2. First, we have $x+y=\sqrt{y^{2}+1}-\sqrt{x^{2}+1}$. Squaring both sides and simplifying, we get
$$
\sqrt{x^{2}+1} \cdot \sqrt{y^{2}+1}=1-x y \text {. }
$$

Squaring both sides again, we get
$$
x^{2} y^{2}+x^{2}+y^{2}+1=x^{2} y^{2}-2 x y+1 \text {, }
$$

which simplifies to $(x+y)^{2}=0$.
Therefore, $x+y=0$.