 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/5/a/ii/solution

\frac{4}{( x - 1 ) ^{2} ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C}{x + 1} .

Multiplying by

(x - 1)^{2} (x + 1)

gives

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

. Setting

x = 1, - 1

gives

B = 2, C = 1

, and comparing the coefficient of

x^{2}

gives

A = - 1

. Thus

f (x) = - \frac{1}{x - 1} + \frac{2}{( x - 1 ) ^{2}} + \frac{1}{x + 1} .

Integrating each term gives the antiderivative

\int f (x) d x = lo g \frac{x + 1}{x - 1} - \frac{2}{x - 1} + C .

The constant of integration can be chosen independently on each interval separated by

- 1

and

1

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