Let $a = \sqrt{x^2 + 2x + 2} > 0$ and $b = \sqrt{x^2 + 1} > 0$. Therefore
$$x = \frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \frac{a^2 - b^2 - 1}{2}$$
$$x + 1 = \frac{a^2 - b^2 + 1}{2}.$$
The equation is equivalent to:
$$\frac{a^2 - b^2 + 1}{2} \cdot a + \frac{a^2 - b^2 - 1}{2} \cdot b = 0$$
$$(a^2 - b^2)a + a + (a^2 - b^2)b - b = 0$$
$$(a - b)((a + b)^2 + 1) = 0.$$
Hence, $a = b$ and $x = -\frac{1}{2}$. $\square$