 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/2/b/solution

u = ln x

, so

d u = d x / x

. The real domain is

x > 1

, ensuring

u > 0

\int \frac{l n ( l n x )}{x} d x = \int ln u d u = u ln u - u + C .

Thus

ln x ln (ln x) - ln x + C, x > 1.

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