Since [tex]x^2-1=(x-1)(x+1)[/tex], by the remainder theorem we have
[tex]\dfrac{p(x)}{x-1} = q(x) + \dfrac{p(1)}{x-1} = q(x) + \dfrac{11}{x-1}[/tex]
where [tex]p(x) = x^{10}+x^9+\cdots+x+1[/tex].
Then
[tex]\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac{q(-1)}{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{q(-1)(x-1) + 11}{x^2-1}[/tex]
The only missing piece is q(x), which we can get through usual polynomial division:
[tex]\dfrac{x^{10}+x^9+\cdots+x+1}{x-1} = \underbrace{x^9 + 2x^8 + 3x^7 + \cdots + 9x + 10}_{q(x)} + \dfrac{11}{x-1}[/tex]
so that
[tex]q(-1) = (-1)^9 + 2(-1)^8 + \cdots + 9(-1) + 10 = 5[/tex]
Then the remainder we want is
[tex]\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac5{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{5(x-1)+11}{x^2-1} = q^*(x) + \dfrac{\boxed{5x+6}}{x^2-1}[/tex]
