Question

решить интеграл dx/4x^2-5

Answer 1

Ответ:

Табличный интеграл :     $\bf \displaystyle \int \frac{du}{u^2-a^2}=\frac{1}{2a}\cdot ln\Big|\, \frac{u-a}{u+a}\, \Big|+C$    .

$\bf \displaystyle \int \frac{dx}{4x^2-5}=\frac{1}{2}\int \frac{d(2x)}{(2x)^2-(\sqrt5)^2}=\frac{1}{2\cdot }\cdot \frac{1}{2\sqrt5}\cdot ln\, \Big|\, \frac{2x-\sqrt5}{2x+\sqrt5}\, \Big|+C=\\\\\\=\frac{1}{4\sqrt5}\cdot ln\, \Big|\, \frac{2x-\sqrt5}{2x+\sqrt5}\, \Big|+C$