Вычислите интеграл [tex]\int\limits {\frac{x^2+7x-2}{x^3+8} } \, dx[/tex]. Created by sobakazabiyaka. algebra-ru

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Miroslava227

Ответ:

${x}^{3} + 8 = (x + 2)( {x}^{2} - 2x + 4)$

раздели на простейшие дроби с помощью неопределенных коэффициентов

$\frac{ {x}^{2} + 7x - 2 }{(x + 2)( {x}^{2} -2x + 4)} = \frac{A}{x + 2} + \frac{Bx + C}{ {x}^{2} - 2x + 4 } \\ {x}^{2} + 7x - 2 = A( {x}^{2} - 2x + 4) + (Bx + C)(x + 2) \\ {x}^{2} + 7x - 2 = A {x}^{2} - 2 Ax + 4A + B {x}^{2} + 2Bx + Cx + 2C \\ \\ 1 = A+ B\\ 7 = - 2A + 2B+ C\\ - 2 = 4A + 2C\\ \\ B = 1 - A \\ C = - 1 - 2A \\ - 2A + 2 - 2A - 1 - 2A = 7 \\ \\ - 6A = 6 \\ A = - 1 \\ \\ B=2 \\ C = 1$

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

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