 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/1/b/solution

The repeated factor requires both a simple-pole and a double-pole term in the partial fraction decomposition. Write

\frac{1}{( x - 1 ) ^{2} ( x + 1 )} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{( x - 1 ) ^{2}} .

Multiplying through gives

1 = A (x - 1)^{2} + B (x^{2} - 1) + C (x + 1)

. Substituting

x = - 1

and

x = 1

gives

A = 1/4

and

C = 1/2

; the quadratic coefficient then gives

B = - 1/4

. Therefore

\frac{1}{x ^{3} - x ^{2} - x + 1} = \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x - 1 )} + \frac{1}{2 ( x - 1 ) ^{2}}, x \neq = \pm 1.

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