Question

Помогите Срочно пж!
найти неопределенный интеграл

Answer 1

Ответ:

$\int\limits \frac{ {x}^{2} - 6x + 8 }{ {x}^{3} + 8 } dx = \int\limits \frac{ {x}^{2} - 6x + 8 }{(x + 2)( {x}^{2} - 2x + 4)} dx$

С помощью неопределенных коэффициентов:

$\frac{ {x}^{2} - 6x + 8}{(x + 2)( {x}^{2} - 2x + 4) } = \frac{a}{x + 2} + \frac{bx + c}{ {x}^{2} - 2x + 4 } \\ x {}^{2} - 6x + 8 = a( {x}^{2} - 2x + 4) + (bx + c)(x + 2) \\ {x}^{2} - 6x + 8 = a {x}^{2} - 2ax + 4a + b {x}^{2} + 2 bx + cx + 2c \\ \\ 1 = a + b \\ - 6 = - 2a + 2b + c \\ 8 = 4a + 2c \\ \\ b = 1 - a \\ c = 4 - 2a \\ - 6 = - 2a + 2 - 2a + 4 - 2a \\ \\ - 6a = - 12 \\ \\ a = 2 \\ b = - 1\\ c = 0$

$\int\limits \frac{2dx}{x + 2} - \int\limits \frac{dx}{ {x}^{2} - 2x + 4} = \\ = 2 ln(x + 2) - \int\limits \frac{dx}{ {x}^{2} - 2x + 1 + 3 } = \\ = 2 ln(x + 2) - \int\limits \frac{d(x - 1)}{ {(x - 1)}^{2} + {( \sqrt{3}) }^{2} } = \\ = 2 ln(x + 2) - \frac{1}{ \sqrt{3} } arctg( \frac{x - 1}{ \sqrt{3} } ) + C$