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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.5.9

5–16. Set up the appropriate form of the partial fraction decomposition for the following expressions. Do not find the values of the unknown constants.
9. 4/(x⁵ - 5x³ + 4x)

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First, factor the denominator completely. The denominator is \(x^{5} - 5x^{3} + 4x\). Start by factoring out the common factor \(x\): \[x^{5} - 5x^{3} + 4x = x(x^{4} - 5x^{2} + 4)\]
Next, factor the quartic polynomial \(x^{4} - 5x^{2} + 4\). Treat \(x^{2}\) as a variable, say \(y = x^{2}\), so the expression becomes \(y^{2} - 5y + 4\). Factor this quadratic: \[y^{2} - 5y + 4 = (y - 4)(y - 1)\] Substitute back \(y = x^{2}\) to get: \[ (x^{2} - 4)(x^{2} - 1) \]
Further factor the difference of squares: \[x^{2} - 4 = (x - 2)(x + 2)\] \[x^{2} - 1 = (x - 1)(x + 1)\] So the full factorization of the denominator is: \[x(x - 2)(x + 2)(x - 1)(x + 1)\]
Since all factors are linear and distinct, the partial fraction decomposition will have terms of the form: \[\frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} + \frac{D}{x - 1} + \frac{E}{x + 1}\] where \(A\), \(B\), \(C\), \(D\), and \(E\) are constants to be determined.
Set up the equation: \[\frac{4}{x^{5} - 5x^{3} + 4x} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} + \frac{D}{x - 1} + \frac{E}{x + 1}\] This is the appropriate form of the partial fraction decomposition without solving for the constants.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions with denominators that are factors of the original denominator. This technique simplifies integration and other operations by breaking down complex expressions into manageable parts.
추천 영상:
가이드 코스
10:07
Partial Fraction Decomposition: Distinct Linear Factors

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its irreducible factors. For partial fractions, factoring the denominator completely into linear and/or irreducible quadratic factors is essential to determine the form of the decomposition.
추천 영상:
07:00
Taylor Polynomials

Form of Partial Fractions for Repeated and Higher-Degree Factors

When the denominator has repeated linear factors or higher-degree factors, the partial fraction decomposition includes terms for each power of the repeated factor and numerators of appropriate degree for irreducible quadratic factors. Setting up the correct form is crucial before solving for unknown constants.
추천 영상:
가이드 코스
11:26
Partial Fraction Decomposition: Repeated Linear Factors