First of all, we have to observe that the expression makes sense only if
[tex]x\neq\pm 1[/tex]
otherwise, you'd have zero denominators, which are not allowed. So, assuming [tex]x\neq\pm 1[/tex], let's sum the two fractions at the left hand side:
[tex]\dfrac{2}{x+1}+\dfrac{x}{x-1} = \dfrac{2(x-1)+x(x+1)}{(x+1)(x-1)} = \dfrac{2x-2+x^2+x}{x^2-1} = \dfrac{x^2+3x-2}{x^2-1}[/tex]
Again, since we're assuming [tex]x\neq\pm 1[/tex], we can multiply both sides by [tex]x^2-1[/tex] to get
[tex]x^2+3x-2=2 \iff x^2+3x-4=0[/tex]
This equation has solutions [tex]x=-4[/tex] and [tex]x=1[/tex], but we can only accept the first, since 1 is a forbidden value.
So, the only solutions to this equation is [tex]x=-4[/tex]
