Question

.надо найти интеграл​

Answer 1

Ответ:

$\displaystyle \int sin(4x+5)\, dx=-\frac{1}{4}\, cos(4x+5)+C\\\\\\\int (2x-4)^{12}\, dx=\frac{1}{2}\cdot \frac{(2x-4)^{13}}{13}+C\\\\\\\int cos(3t-1)\, dt=\frac{1}{3}\, sin(3t-1)+C\\\\\\\int \frac{4}{cos^2(2x+1)}\, dx=4\cdot \frac{1}{2}\, tg(2x+1)+C=2\, tg(2x+1)+C\\\\\\\int \frac{dx}{(6x-9)^4}=\frac{1}{6}\cdot \frac{(6x-9)^{-3}}{-3}+C=-\frac{1}{18\, (6x-9)^3}+C\\\\\\\int \frac{5\, dx}{1+(9x-2)}=5\cdot \int \frac{dx}{9x-1}=\frac{5}{9}\cdot ln|9x-1|+C$

$\star \ \ \displaystyle \int \frac{5\, dx}{1+(9x-2)^2}=\frac{5}{9}\cdot arctg(9x-2)+C\ \ \star \\\\\\\star \ \ \int f(x)\, dx=F(x)+C\ \ \Rightarrow \ \ \int f(kx+b)\, dx=\frac{1}{k}\cdot F(kx+b)+C\ \ \star$