 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/17/a/solution

The original PDF places the

- 1

outside the square root. Use the substitution rule with

t = x

, so

d x = 2 t d t

and the endpoints become

2

and

3

\int_{2}^{3} \frac{2 t}{t - 1} d t = 2 \int_{2}^{3} (1 + \frac{1}{t - 1}) d t = 2 [t + ln (t - 1)]_{2}^{3} .

2 + 2 ln 2.

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