Question

Помогите решить !!!!​

Answer 1

Ответ:

Решается путём разложения на простейшие дроби.

1.

$\int\limits \frac{3x + 8}{(x - 2)(x + 5)} dx \\ \\ \frac{3x + 8}{(x - 2)(x + 5)} = \frac{a}{x - 2} + \frac{b}{x + 5} \\ 3x + 8 = a(x + 5) + b(x - 2) \\ 3x + 8 = ax + 5a + bx - 2b \\ \\ 3 = a + b \\ 8 = 5a - 2b \\ \\ a = 3 - b \\ 8 = 15 - 5b - 2b \\ \\ - 7b = - 7\\ b = 1 \\ \\ a = 3 - 1 = 2 \\ \\ \int\limits \frac{2dx}{x - 2} + \int\limits \frac{1 \times dx}{x + 5} = \\ = 2\int\limits \frac{d(x - 2)}{x - 2} + \int\limits \frac{d(x + 5)}{x + 5} = \\ = 2 ln( |x - 2| ) + ln( |x + 5| ) + C$

2.

$\int\limits \frac{7x + 12}{(x - 1)(3x + 1)} dx \\ \\ \frac{7x + 12}{(x - 1)(3x + 1)} = \frac{a}{x - 1} + \frac{b}{3x + 1} \\ 7x + 12 =a (3x + 1) + b(x - 1) \\ 7x + 12 = 3ax + a + bx - b \\ \\ 7 = 3a + b \\ 12 = a - b \\ (1) + (2) \\ \\ 19 = 4a \\ a = \frac{19}{4} \\ b = 7 - 3a = 7 - \frac{57}{4} = - \frac{29}{19}$

$\frac{19}{4} \int\limits \frac{dx}{x - 1} - \frac{29}{4} \int\limits \frac{dx}{3x + 1} = \\ = \frac{19}{4} ln( |x - 1| ) - \frac{29}{4} \times \frac{1}{3} \int\limits \frac{d(3x + 1)}{3x + 1} = \\ = \frac{19}{4} ln( |x - 1| ) - \frac{29}{12} ln( |3x + 1| ) + C$