Question

Срочно! Пожалуйста!
найти интегралы от рациональных функций.

5)

Answer 1

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

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