 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/17/c/i/solution

u = 1

, the integrand is unbounded but integrable:

1/ u^{2} - 1 \sim [2 (u - 1)]^{- 1/2}

. Thus the lower-limit integral is understood as an improper integral. With

z = u + u^{2} - 1

\frac{d z}{d u} = 1 + \frac{u}{u ^{2} - 1} = \frac{z}{u ^{2} - 1}, \frac{d u}{u ^{2} - 1} = \frac{d z}{z} .

The lower endpoint corresponds to

z = 1

F (x) = ln (x + x^{2} - 1) = arcosh x, x > 1.

The limiting value at

x ↓ 1

is zero. Here

arcosh

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