 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/17/c/solution

Take

t = e^{x} > 0

d x = d t / t

. The transformed integral is

\int \frac{t - 2}{t ^{2} + 1} d t = \frac{1}{2} ln (t^{2} + 1) - 2 arctan t + C .

The two terms follow from the derivatives of the natural logarithm and the inverse tangent. Returning to

x

\frac{1}{2} ln (1 + e^{2 x}) - 2 arctan (e^{x}) + C .

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