Q. Simplify \( \frac{x^3 + x^2 + x + 2}{x^2 – 1} \).

Step-by-step solution

1. Divide the polynomial \(x^3 + x^2 + x + 2\) by \(x^2 – 1\) using polynomial long division.
   • First term: divide \(x^3\) by \(x^2\) to get \(x\).
   • Multiply divisor by \(x\): \(x(x^2 – 1) = x^3 – x\).
   • Subtract: \((x^3 + x^2 + x + 2) – (x^3 – x) = x^2 + 2x + 2\).
2. Continue the division.
   • Divide \(x^2\) by \(x^2\) to get \(1\).
   • Multiply divisor by \(1\): \(1(x^2 – 1) = x^2 – 1\).
   • Subtract: \((x^2 + 2x + 2) – (x^2 – 1) = 2x + 3\).
3. Write the quotient and remainder form.

   \(x^3 + x^2 + x + 2\) divided by \(x^2 – 1\) equals \(x + 1 + \frac{2x + 3}{x^2 – 1}\).

4. Factor the denominator for partial fractions: \(x^2 – 1 = (x – 1)(x + 1)\). Decompose the remainder fraction:
   \[ \frac{2x+3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \]
   Multiply through by \((x-1)(x+1)\) to get:
   \[ 2x + 3 = A(x+1) + B(x-1) \]
   Equate coefficients:

   \(A + B = 2\) (coefficient of \(x\)) and \(A – B = 3\) (constant term).

   Solve the system:

   \(A = \frac{5}{2}\) and \(B = -\frac{1}{2}\).

5. Substitute the partial fractions back:
   \[ \frac{2x+3}{x^2-1} = \frac{5/2}{x-1} – \frac{1/2}{x+1} \]
   Therefore the full simplified form is:

   \(x + 1 + \dfrac{5/2}{x-1} – \dfrac{1/2}{x+1}\).