 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/2/b/solution

u = ln x

and

d v = x d x

. Since

d u = d x / x

and

v = x^{2} /2

\int x ln x d x = \frac{x ^{2}}{2} ln x - \frac{1}{2} \int x d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + C, x > 0.

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