Question

Помогите решить интеграл

Answer 1

Ответ:

$\int\limits \frac{ {x}^{2} + 3x + 2 }{ {x}^{3} - 1 } dx = \int\limits \frac{ {x}^{2} + 3x + 2}{(x - 1)( {x}^{2} + x + 1)}$

Разделим на простейшие дроби:

$\frac{ {x}^{2} + 3x + 2 }{(x - 1)( {x}^{2} + x + 1) } = \frac{a}{x - 1} + \frac{bx + c}{ {x}^{2} + x + 1} \\ {x}^{2} + 3x + 2 = a( {x}^{2} + x + 1) + (bx + c)(x - 1) \\ {x}^{2} + 3x + 2 = a {x}^{2} + ax + a + b {x}^{2} - bx + cx - c \\ \\ 1 = a + b \\ 3 = a - b + c \\ 2 = a - c \\ \\ b = 1 - a \\ c = a - 2\\ \\ \\ 3 = a - 1 + a + a - 2 \\ 3a = 6 \\ a = 2 \\ \\ b = - 1 \\ c = 0$

Получаем:

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