Step 1: Factor the denominator completely. The given expression is: $$ \frac{x^3 - x^2 - 3x + 5}{(x-1)(x^2-1)} $$ We know that $x^2-1 = (x-1)(x+1)$. Substitute this into the denominator: $$ (x-1)(x^2-1) = (x-1)(x-1)(x+1) = (x-1)^2(x+1) $$ So the expression becomes: $$ \frac{x^3 - x^2 - 3x + 5}{(x-1)^2(x+1)} $$ Step 2: Check if the fraction is proper. The degree of the numerator ($x^3 - x^2 - 3x + 5$) is 3. The degree of the denominator ($(x-1)^2(x+1) = (x^2-2x+1)(x+1) = x^3 - x^2 - x + 1$) is 3. Since the degree of the numerator is equal to the degree of the denominator, this is an improper rational fraction. We must perform polynomial long division first. Step 3: Perform polynomial long division. Divide $x^3 - x^2 - 3x + 5$ by $x^3 - x^2 - x + 1$. $$ \frac{x^3 - x^2 - 3x + 5}{x^3 - x^2 - x + 1} = 1 + \frac{(x^3 - x^2 - 3x + 5) - (x^3 - x^2 - x + 1)}{x^3 - x^2 - x + 1} $$ $$ = 1 + \frac{-2x + 4}{x^3 - x^2 - x + 1} $$ So, the expression can be written as: $$ 1 + \frac{4 - 2x}{(x-1)^2(x+1)} $$ Step 4: Set up the partial fraction decomposition for the proper fraction. Let $f(x) = \frac{4 - 2x}{(x-1)^2(x+1)}$. The denominator has a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+1)$. The form of the partial fraction decomposition is: $$ \frac{4 - 2x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} $$ Step 5: Clear the denominators to solve for A, B, and C. Multiply both sides by $(x-1)^2(x+1)$: $$ 4 - 2x = A(x-1)(x+1) + B(x+1) + C(x-1)^2 $$ $$ 4 - 2x = A(x^2-1) + B(x+1) + C(x^2-2x+1) $$ Step 6: Solve for A, B, and C by substituting convenient values for $x$. • Let $x=1$: $$ 4 - 2(1) = A(1^2-1) + B(1+1) + C(1-1)^2 $$ $$ 2 = A(0) + B(2) + C(0) $$ $$ 2 = 2B \implies B = 1 $$ • Let $x=-1$: $$ 4 - 2(-1) = A((-1)^2-1) + B(-1+1) + C(-1-1)^2 $$ $$ 4 + 2 = A(0) + B(0) + C(-2)^2 $$ $$ 6 = 4C \implies C = \frac{6}{4} = \frac{3}{2} $$ • Let $x=0$: $$ 4 - 2(0) = A(0^2-1) + B(0+1) + C(0-1)^2 $$ $$ 4 = A(-1) + B(1) + C(1) $$ $$ 4 = -A + B + C $$ Substitute the values of $B=1$ and $C=\frac{3}{2}$: $$ 4 = -A + 1 + \frac{3}{2} $$ $$ 4 = -A + \frac{2}{2} + \frac{3}{2} $$ $$ 4 = -A + \frac{5}{2} $$ $$ A = \frac{5}{2} - 4 $$ $$ A = \frac{5}{2} - \frac{8}{2} $$ $$ A = -\frac{3}{2} $$ Step 7: Write the complete partial fraction decomposition. Substitute the values of A, B, and C back into the expression from Step 3: $$ 1 + \frac{-\frac{3}{2}}{x-1} + \frac{1}{(x-1)^2} + \frac{\frac{3}{2}}{x+1} $$ $$ 1 - \frac{3}{2(x-1)} + \frac{1}{(x-1)^2} + \frac{3}{2(x+1)} $$ The final answer is $\boxed{1 - \frac{3}{2(x-1)} + \frac{1}{(x-1)^2} + \frac{3}{2(x+1)}}$.