[tex]\displaystyle \frac{4x^2+x-3}{x^2-1}=\frac{4x^2+4x-3x-3}{(x-1)(x+1)}=\frac{4x(x+1)-3(x+1)}{(x-1)(x+1)}=\frac{(x+1)(4x-3)}{(x-1)(x+1)}=\\\frac{4x-3}{x-1}[/tex]