Question

интеграл

((5x-7)/(x^2-x-20))dx

Answer 1

$\displaystyle \int\frac{5x-7}{x^2-x-20}dx=\int\frac{5x-7}{(x+4)(x-5)}=\\\\\\=3\int\frac{d(x+4)}{x+4}+2\int\frac{d(x-5)}{x-5}=3ln|x+4|+2ln|x-5|+C\\\\\\\frac{5x-7}{(x+4)(x-5)}=\frac{A}{x+4}+\frac{B}{x-5}=\frac{3}{x+4}+\frac{2}{x-5}\\5x-7=A(x-5)+B(x+4)\\x|5=A+B\\x^0|-7=-5A+4B\\18=9B\\B=2;A=3$
Проверка:
$(3ln|x+4|+2ln|x-5|+C)'=\frac{3}{x+4}+\frac{2}{x-5}=\frac{3(x-5)+2(x+4)}{(x+4)(x-5)}=\frac{5x-7}{x^2-x-20}$