Question

Найти интеграл l dx/ 4x^2 - 9

Answer 1

Ответ:

$\int\limits \frac{dx}{4 {x}^{2} - 9 } = \int\limits \frac{dx}{(2x) {}^{2} - {3}^{2} } = \frac{1}{2} \int\limits \frac{d(2x)}{(2x) {}^{2} - {3}^{2} } = \\ = \frac{1}{2} \times \frac{1}{2 \times 3} ln( | \frac{2x - 3}{2x + 3} | ) + C = \frac{1}{12} ln( | \frac{2x - 3}{2x + 3} | ) + cC$