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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.5.16

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (x + 3) / (2x³ - 8x) dx

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First, factor the denominator to simplify the integrand. The denominator is \(2x^{3} - 8x\). Factor out the common term \$2x\( to get \(2x(x^{2} - 4)\), and then recognize that \)x^{2} - 4$ is a difference of squares, so it factors further as \(2x(x - 2)(x + 2)\).
Rewrite the integrand as a sum of partial fractions with unknown coefficients: \(\frac{x + 3}{2x(x - 2)(x + 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}\). Here, \(A\), \(B\), and \(C\) are constants to be determined.
Multiply both sides of the equation by the denominator \(2x(x - 2)(x + 2)\) to clear the fractions, resulting in: \(x + 3 = A \cdot 2(x - 2)(x + 2) + B \cdot 2x(x + 2) + C \cdot 2x(x - 2)\).
Expand the right-hand side and collect like terms in powers of \(x\). Then, equate the coefficients of corresponding powers of \(x\) on both sides to form a system of equations for \(A\), \(B\), and \(C\).
Solve the system of equations to find the values of \(A\), \(B\), and \(C\). Once found, rewrite the integral as the sum of simpler integrals: \(\int \frac{A}{x} dx + \int \frac{B}{x - 2} dx + \int \frac{C}{x + 2} dx\), and then integrate each term separately.

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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 technique used to break down a complex rational function into simpler fractions that are easier to integrate. It involves expressing the integrand as a sum of fractions with simpler denominators, typically linear or quadratic factors. This method is essential when integrating rational functions where the degree of the numerator is less than the degree of the denominator.
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Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of its factors, such as linear or quadratic expressions. For partial fractions, factoring the denominator completely is crucial to identify the terms in the decomposition. In this problem, factoring the cubic denominator helps determine the form of the partial fractions needed for integration.
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Integration of Rational Functions

Integration of rational functions often requires rewriting the integrand into simpler parts, such as partial fractions, to apply basic integration rules. After decomposition, each term can be integrated using standard formulas, including logarithmic and inverse trigonometric integrals. Understanding these integration techniques is key to solving the integral effectively.
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