Question

Помогите найти первообразную!

Answer 1

$\displaystyle 1) \int -3dx=-3x+C$
$\displaystyle 2) \int 2x-4= \frac{2x^2}{2}-4=x^2-4x+C$
$\displaystyle 3) \int (6x^5+8x)dx=x^6+4x^2+C$
$\displaystyle 4) \int (x+9)dx= \frac{x^2}{2}+9x+C$
$\displaystyle 5) \int \frac{1}{x^3}-12x^{11}dx=\int x^{-3}-12x^{11} dx =- \frac{1}{2x^2}-x^{12}+C$
$\displaystyle 6) \int 7(4-7x)^6dx=7*(- \frac{1}{7}) \int (4-7x)^6dx=- \frac{1}{7}* \frac{7(4-7x)^7}{7}=$
$\displaystyle =- \frac{(4-7x)^7}{7}+C$
$\displaystyle 7) \int 2(4x+1)dx=2*( \frac{4x^2}{2}+x)=4x^2+2x=2x(2x+1)+C$
$\displaystyle 8) \int sin(5x-7)dx=- \frac{1}{5}cos(5x-7)=- \frac{cos(5x-7)}{5}+C$
$\displaystyle 9) \int \frac{4}{(9x+3)^4} dx=4* \frac{1}{9} \int \frac{1}{(9x+3)^4}dx= \frac{4}{9}*(- \frac{1}{3(9x+3)^3})=$
$\displaystyle =- \frac{4}{27(9x+3)^3}+C$
$\displaystyle 10) \int \frac{3}{cos^25x}dx=3* \frac{1}{5} \int \frac{1}{cos^25x} dx = \frac{3tg5x}{5} +C$