Question

Найдите интеграл.

Answer 1

$\int\limits {\sqrt{4x-5}} \, dx = \frac{1}4\int\limits {(4x-5)^{\frac{1}2}} \, d(4x) =\frac{1}4*\frac{(4x-5)^{\frac{1}2+1}}{\frac{1}2+1}+C=\\=\frac{1}{4}*\frac{2}{3}(4x-5)^\frac{3}2+C=\frac{1}6\sqrt{(4x-5)^3}+C$

$\int\limits {cos(4x-2)} \, dx=\frac{1}4\int\limits {cos(4x-2)} \, d(4x) =\frac{1}4sin(4x-2)+C$