15. For a real number $a$, there is only one real value of $x$ that satisfies the equation
$$
\frac{x+1}{x-1}+\frac{x-1}{x+1}+\frac{2 x+a+2}{x^{2}-1}=0 \text {. }
$$

Find the sum of all such real numbers $a$.
Three, 15. The equation can be transformed into
$$
2 x^{2}+2 x+a+4=0 .
$$

When equation (1) has two equal real roots, we have
$$
\Delta=4-4 \times 2 \times(a+4)=0 \text {. }
$$

From this, we get $a_{1}=-\frac{7}{2}$.
At this point, (1) has one root $x=-\frac{1}{2}$.
Verification shows that $x=-\frac{1}{2}$ satisfies the equation in the problem.
When equation (1) has two distinct real roots, we have
$$
\Delta=4-4 \times 2 \times(a+4)>0 \text {. }
$$

From this, we get $a<-\frac{7}{2}$.
If $x=1$ is a root of equation (1), then the original equation has an extraneous root $x=1$; substituting into (1) yields $a_{2}=-8$. At this point, the other root of (1) is $x=-2$, which also satisfies the equation in the problem.

If $x=-1$ is a root of equation (1), then the original equation has an extraneous root $x=-1$, substituting into (1) yields $a_{3}=-4$. At this point, the other root of (1) is $x=0$. Verification shows that $x=0$ also satisfies the equation in the problem.
Therefore, $a_{1}=-\frac{7}{2}, a_{2}=-8, a_{3}=-4$ are the solutions. Hence, $a_{1}+a_{2}+a_{3}=-\frac{31}{2}$.