 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/17/a/ii/solution

u = ln x

and

d v = x^{- 4} d x

. Then

d u = d x / x

and

v = - 1/ (3 x^{3})

\int \frac{l n x}{x ^{4}} d x = - \frac{l n x}{3 x ^{3}} + \frac{1}{3} \int x^{- 4} d x = - \frac{3 ln x + 1}{9 x ^{3}} + C, x > 0.

Here

C

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