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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.5.9
Chapter 8, Problem 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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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.
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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.
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Partial Fraction Decomposition: Repeated Linear Factors
Related Practice
Textbook Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

48. ∫ √(9 - 4x²) dx

Textbook Question

Let f(x) = (4x³ + x² + 4x + 2) / (x² + 1). Use long division to show that f(x) = 4x + 1 + 1 / (x² + 1) and use this result to evaluate ∫f(x) dx.

Textbook Question

1. On which derivative rule is integration by parts based?

Textbook Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

20. ∫ sin⁻¹(x) dx

Textbook Question

7–64. Integration review Evaluate the following integrals.

20. ∫ eˣ (1 + eˣ)⁹ (1 - eˣ) dx

Textbook Question

River flow rates

The following figure shows the discharge rate r(t) of the Snoqualmie River near Carnation, Washington, starting on a February day when the air temperature was rising. The variable t is the number of hours after midnight, r(t) is given in millions of cubic feet per hour, and ∫(0 to 24) r(t) dt equals the total amount of water that flows by the town of Carnation over a 24-hour period. Estimate ∫(0 to 24) r(t) dt using the Trapezoidal Rule and Simpson's Rule with the following values of n.


n = 6