 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/13/a/ii/solution

Let

D = 3 x^{2} + 2 x + 2

. Splitting the numerator into a derivative term and a constant gives

x - 1 = D^{'} /6 - 4/3

. Thus

\int \frac{x - 1}{D} d x = \frac{1}{6} lo g D - \frac{4}{3} \int \frac{d x}{D} .

Complete the square as

D = ((3 x + 1)^{2} + 5) /3

and substitute

u = 3 x + 1

. The remaining integral is

\int d u / (u^{2} + 5) = arctan (u / 5) / 5

. Hence

\frac{1}{6} lo g (3 x^{2} + 2 x + 2) - \frac{4}{3 5} arctan \frac{3 x + 1}{5} + C .

The quadratic denominator is positive for all real

x

, so no logarithmic absolute value or real-domain exclusion is needed.

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