Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$ The partial fraction of  $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.
I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?

Hint. you may write
$$
\frac{x^2+2}{x^2-1}=\frac{(x^2-1)+3}{x^2-1}=\frac{x^2-1}{x^2-1}+\frac{3}{x^2-1}=\color{red}{1+}\frac{3}{x^2-1}
$$
