 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/a/ii/solution

u = arctan x

and

d v = x^{- 2} d x

. Then

d u = d x / (1 + x^{2})

and

v = - 1/ x

, so

\int \frac{a r c t a n x}{x ^{2}} d x = - \frac{a r c t a n x}{x} + \int \frac{d x}{x ( 1 + x ^{2} )} .

1/ [x (1 + x^{2})] = 1/ x - x / (1 + x^{2})

. Therefore

\int \frac{arctan x}{x ^{2}} d x = - \frac{arctan x}{x} + ln ∣ x ∣ - \frac{1}{2} ln (1 + x^{2}) + C, x \neq = 0.

The constant of integration may be chosen independently on the positive and negative intervals.

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