Answer
The solution set of the equation is all real numbers except $x=1\text{ and }x=-1$.
Work Step by Step
Consider the equation $\frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{{{x}^{2}}-1}$.
The equation is not defined for $x=1\text{ and }x=-1$.
First of all, simplify the equation.
That is,
$\begin{align}
& \frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{{{x}^{2}}-1} \\
& \frac{\left( x+1 \right)-\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)}=\frac{2}{{{x}^{2}}-1} \\
& \frac{x+1-x+1}{{{x}^{2}}-1}=\frac{2}{{{x}^{2}}-1} \\
& \frac{2}{{{x}^{2}}-1}=\frac{2}{{{x}^{2}}-1}
\end{align}$
Now, multiply both sides by ${{x}^{2}}-1$ and get,
$\begin{align}
& \frac{2}{{{x}^{2}}-1}\cdot \left( {{x}^{2}}-1 \right)=\frac{2}{{{x}^{2}}-1}\cdot \left( {{x}^{2}}-1 \right) \\
& 2=2
\end{align}$
Therefore, all the real numbers except $x=1\text{ and }x=-1$ are solutions for the equation.
Hence, the solution set of the equation is all real numbers except $x=1\text{ and }x=-1$.
