Answer
$\dfrac{3x^2-2x-3}{(x-1)(x+1)}; x\ne -1, 1$
Work Step by Step
The LCM of $x+1$ and $x-1$ is $(x+1)(x-1)$.
Use the LCM as the expressions' LCD to make them similar:
$=\dfrac{x\color{blue}{(x-1)}}{(x+1)\color{blue}{(x-1)}} +\dfrac{(2x-3)\color{blue}{(x+1)}}{(x-1)\color{blue}{(x+1)}}
\\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x(x+1)-3(x+1)}{(x-1)(x+1)}
\\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x^2+2x-3x-3}{(x-1)(x+1)}
\\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x^2-x-3}{(x-1)(x+1)}$
The expressions are similar so add the numerators and copy the denominator to obtain:
$=\dfrac{x^2-x+2x^2-x-3}{(x-1)(x+1)}
\\=\dfrac{3x^2-2x-3}{(x-1)(x+1)}; x\ne -1, 1$
