Right away you know that you can't have [tex]x=\pm1[/tex] as possible solutions because either of these make the first term's denominator 0, while [tex]x=-1[/tex] also makes the second term's denominator 0.
With this in mind, let's fix [tex]x\neq\pm1[/tex], which allows you to write
[tex]\dfrac8{x^2-1}+\dfrac4{x+1}=1\iff8+4(x-1)=x^2-1[/tex]
Moving everything to one side, you have
[tex]x^2-4x-5=0\iff (x-5)(x+1)=0\implies x=5\text{ or }x=-1[/tex]
We omitted [tex]x=-1[/tex] from the solution set, so this is an extraneous root and we're left with a solution of [tex]x=5[/tex].
