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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.5.12
Chapter 8, Problem 8.5.12

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (2x + 1) / (x² - 7x + 12) dx

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First, factor the quadratic denominator \(x^{2} - 7x + 12\). Find two numbers that multiply to 12 and add to -7. This factors as \((x - 3)(x - 4)\).
Express the integrand as a sum of partial fractions: write \(\frac{2x + 1}{(x - 3)(x - 4)} = \frac{A}{x - 3} + \frac{B}{x - 4}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the denominator \((x - 3)(x - 4)\) to clear the fractions: \(2x + 1 = A(x - 4) + B(x - 3)\).
Expand the right side and collect like terms: \(2x + 1 = A x - 4A + B x - 3B = (A + B) x + (-4A - 3B)\). Then, equate the coefficients of \(x\) and the constant terms on both sides to form a system of equations: \(2 = A + B\) and \(1 = -4A - 3B\).
Solve the system of equations for \(A\) and \(B\). Once you find \(A\) and \(B\), rewrite the integral as \(\int \frac{A}{x - 3} dx + \int \frac{B}{x - 4} dx\) and integrate each term separately using the natural logarithm rule for integrals of the form \(\int \frac{1}{x - c} dx\).

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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 irreducible quadratic factors.
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Partial Fraction Decomposition: Distinct Linear Factors

Factoring Quadratic Expressions

Factoring quadratic expressions involves rewriting a quadratic polynomial as a product of two binomials. This step is essential for partial fraction decomposition because it identifies the simpler linear factors in the denominator needed to split the integrand.
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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. Once decomposed, each term can be integrated using standard formulas, such as the integral of 1/(x - a) being ln|x - a| + C.
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Related Practice
Textbook Question

In Exercises 17–20, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (x² dx) / ((x - 1)(x² + 2x + 1))

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Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 0 to 1 of (dt / (t - sin t))

(Hint: t ≥ sin t for t ≥ 0)

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Textbook Question

Average Value: Find the average value of the function f(x) = 1 / (1 - sin θ) on the interval [0, π/6].

Textbook Question

Evaluate the integrals in Exercises 33–52.

∫ cot⁶(2x) dx

Textbook Question

In Exercises 67–73, use integration by parts to establish the reduction formula.

∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx

Textbook Question

Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.

∫ cos²(2θ) sin(θ) dθ