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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
Chapter 8, Problem 8.4.62

60–69. Completing the square Evaluate the following integrals.
62. ∫ du / (2u² - 12u + 36)

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Start by examining the quadratic expression in the denominator: \$2u^{2} - 12u + 36\(. To simplify the integral, factor out the coefficient of \)u^{2}$ from the quadratic terms.
Rewrite the denominator as \$2(u^{2} - 6u + 18)$ by factoring out the 2. This will make completing the square inside the parentheses easier.
Complete the square for the expression inside the parentheses: \(u^{2} - 6u + 18\). Recall that to complete the square for \(u^{2} - 6u\), add and subtract \((\frac{6}{2})^{2} = 9\) inside the expression.
Rewrite \(u^{2} - 6u + 18\) as \((u - 3)^{2} + (18 - 9)\), which simplifies to \((u - 3)^{2} + 9\). Now the denominator is \$2[(u - 3)^{2} + 9]$.
Substitute \(w = u - 3\) to simplify the integral to the form \(\int \frac{du}{2(w^{2} + 9)}\). Then, factor out constants and use the standard integral formula for \(\int \frac{dx}{x^{2} + a^{2}}\) to proceed with integration.

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Key Concepts

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

Completing the Square

Completing the square is a technique used to rewrite a quadratic expression in the form ax² + bx + c as a perfect square plus or minus a constant. This simplifies integration by transforming the denominator into a recognizable form, such as (u - h)² + k, which is easier to integrate using standard formulas.
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Integration of Rational Functions

Integrating rational functions often involves manipulating the integrand into a simpler form. When the denominator is a quadratic expression, rewriting it by completing the square allows the use of standard integral formulas involving arctangent or logarithmic functions, depending on the form of the quadratic.
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Standard Integral Formulas Involving Quadratics

Certain integrals involving quadratics have standard results, such as ∫ dx / (x² + a²) = (1/a) arctan(x/a) + C. Recognizing the denominator as a sum of squares after completing the square enables direct application of these formulas, facilitating the evaluation of the integral.
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