Question

Срочно! Пожалуйста!
найти интегралы от рациональных функций.

5)

Answer 1

$\displaystyle\int\frac{2x+5}{x^3-9x^2+14x}dx=\frac{5}{14}\int\frac{dx}{x}-\frac{9}{10}\int\frac{dx}{x-2}+\frac{19}{35}\int\frac{dx}{x-7}=\\=\frac{5}{14}ln|x|-\frac{9}{10}ln|x-2|+\frac{19}{35}ln|x-7|+C\\\\\\\frac{2x+5}{x^3-9x^2+14x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-7}=\frac{5}{14x}+\frac{-9}{10(x-2)}+\frac{19}{35(x-7)}\\2x+5=A(x^2-9x+14)+B(x^2-7x)+C(x^2-2x)\\x^2|0=A+B+C\\x|2=-9A-7B-2C\\x^0|5=14A\to A=\frac{5}{14}\\B=-\frac{9}{10}\ ;C=\frac{19}{35}$

$\displaystyle\int\frac{2x^4-6x^3+20x^2-33x+49}{(x-3)(x^2+4)}dx=\\=\int2xdx+10\int\frac{d(x-3)}{x-3}+\int\frac{d(x^2+4)}{x^2+4}-3\int\frac{dx}{x^2+4}=\\=x^2+10ln|x-3|+ln|x^2+4|-\frac{3}{2}arctg\frac{x}{2}+C\\\\\\\frac{2x^4-6x^3+20x^2-33x+49}{x^3-3x^2+4x-12}=2x+\frac{12x^2-9x+49}{(x-3)(x^2+4)}\\\frac{12x^2-9x+49}{(x-3)(x^2+4)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+4}=\frac{10}{x-3}+\frac{2x-3}{x^2+4}\\12x^2-9x+49=A(x^2+4)+B(x^2-3x)+C(x-3)\\x^2|12=A+B\\x|-9=-3B+C\\x^0|49=4A-3C\\A=10;B=2;C=-3$