Question

Найти интеграл :
4)dx/4x^2+49
5)cos(7x+6)dx
6)dx/16x^2-4

Answer 1

$\displaystyle\int\frac{dx}{4x^2+49}=\frac{1}{14}\int\frac{d(\frac{2x}{7})}{(\frac{2x}{7})^2+1}=\frac{1}{14}arctg\frac{2x}{7}+C$

$\displaystyle\int cos(7x+6)dx=\frac{1}{7}\int cos(7x+6)d(7x+6)=\frac{1}{7}sin(7x+6)+C$

$\displaystyle\int\frac{dx}{16x^2-4}=\frac{1}{8}\int\frac{d(2x)}{4x^2-1}=\frac{1}{16}ln|\frac{2x-1}{2x+1}|+C$