 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/17/b/solution

u = ln x

d u = d x / x

. The bounds become

3

and

4

, and the remaining denominator factorizes as

(u - 1) (u - 2)

\frac{3 u - 4}{( u - 1 ) ( u - 2 )} = \frac{1}{u - 1} + \frac{2}{u - 2} .

Consequently the definite integral is

[ln (u - 1) + 2 ln (u - 2)]_{3}^{4} = ln \frac{3}{2} + 2 ln 2 = ln 6.

All logarithm arguments here are positive on the integration interval.

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