Let $\frac{3x+5}{x^3-x^2-x+1} = \frac{3x+5}{(x^2-1)(x+1)}$ $= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$ or, 3x + 5 = A(x – 1) (x + 1) + B(x + 1) + C(x – 1)2 ...(i)Putting x = 0 in (i), we get, 5 = –A + B + C ...(ii)Putting x = 1 in (i), we get, 3 + 5 = 2B ⇒ B $= \frac{8}{2} = 4$ Putting x = – 1 in (i), we get, $-3 + 5 = C(-1-1)^2$ $\Rightarrow C = \frac{2}{4} = \frac{1}{2}$Now putting the values of B and C in (ii), we get $5 = -A + 4 + \frac{1}{2}$ $\Rightarrow A = \frac{9}{2} - 5 = -\frac{1}{2}$ $= \frac{1}{2} \log \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + C$