Question

Помогите пожалуйста, срочно надо, 85 баллов

Answer 1

Ответ:

$\int\limits^{ 1 } _ { - 2} \frac{dx}{ {x}^{2} + 4x + 13} \\ \\ {x}^{2} + 4x + 13 = {x}^{2} + 4x + 4 + 9 = \\ = (x + 2) {}^{2} + {3}^{2} \\ \\ \int\limits^{ 1 } _ { - 2} \frac{dx}{ {(x + 2)}^{2} + {3}^{2} } = \int\limits^{ 1} _ { - 2} \frac{d(x + 2)}{ {(x + 2)}^{2} + {3}^{2} } = \\ = \frac{1}{3} arctg (\frac{x + 2}{3}) | ^{ 1} _ { - 2} = \\ = \frac{1}{3} (arctg(1) - arctg(0)) = \\ = \frac{1}{3} \times \frac{\pi}{4} = \frac{\pi}{12}$