Question

Найти неопределенный интеграл

$\frac{3x^3+5x^2-4x+28}{x^4-16}$

Answer 1

$\int\limits \frac{3 {x}^{3} + 5 {x}^{2} - 4x + 28}{ {x}^{4} - 16 } = \int\limits \frac{3 {x}^{3} + 5 {x}^{2} - 4x + 28}{( {x}^{2} - 4)( {x}^{2} + 4)} = \\ = \int\limits \frac{3 {x}^{3} + 5 {x}^{2} - 4x + 28}{(x - 2)(x + 2)( {x}^{2} + 4)}$

Раскладываем на простейшие дроби с неопределенных коэффициентов:

$\frac{3 {x}^{3} + 5 {x}^{2} - 4x + 28}{(x - 2)(x + 2)( {x}^{2} + 4) } = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{Cx + D}{ {x}^{2} + 4 } \\ 3 {x}^{3} + 5 {x}^{2} - 4x + 28 = A(x + 2)( {x}^{2} + 4) + B(x - 2)( {x}^{2} + 4) + (Cx + D) ( {x}^{2} - 4) \\ 3 {x}^{3} + 5 {x}^{2} - 4x + 28 = A {x}^{3} + 4Ax + 2A {x}^{2} + 8A + B {x}^{3} + 4Bx - 2B {x}^{2} - 8B + C {x}^{3} - 4Cx + D {x}^{2} - 4D$

система:

$3 =A + B + C \\ 5 = 2A - 2B + D \\ - 4 = 4A + 4B- 4C \\ 28 = 8A - 8B+ 4D\\ \\ A = 2 \\ B = - 1 \\ C =2 \\ D = - 1$

получаем:

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

Answer 2

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