Срочно!!!¡ интеграл

Решите задачу:

$\int \frac{(x+8)dx}{(x+2)(x+3)(x-1)} =Q\\\\ \frac{x+8}{(x+2)(x+3)(x-1)}= \frac{A}{x+2}+ \frac{B}{x+3}+\frac{C}{x-1}\\\\x=-2:\; \; A= \frac{6}{1\cdot (-3)}=-2\\\\x=-3:\; \; B= \frac{5}{-1\cdot (-4)} = \frac{5}{4}\\\\x=1:\; \; C= \frac{9}{3\cdot 4}= \frac{3}{4}\\\\Q=-2\int \frac{dx}{x+2}+\frac{5}{4} \int \frac{dx}{x+3}+ \frac{3}{4}\int \frac{dx}{x-1}= \\\\=-2\, ln|x+2|+\frac{5}{4}\, ln|x+3|+\frac{3}{4}\, ln|x-1|+C$