Answer
Solution set = $\emptyset$.
(no solution)
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-1,+1 \}\qquad (*)$
The last denominator $x^{2}-1$ is factored as $(x-1)(x+1)$
Multiply the equation with the LCD=$(x-1)(x+1)$
$ 1(x+1)+1(x-1)=2\qquad$ ... simplify (distribute)
$x+1+x-1=2$
$2x=2 $
$ x=1 \qquad$ ...Checking (*), this is not a valid solution
Solution set = $\emptyset$.
(no solution)
