Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 515: 22

Answer

$\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{{x - 6}}{{{x^2} - 4}} + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{3{x^2} + 12x - 20}}{{{x^4} - 8{x^2} + 16}}} dx \cr & {\text{Factoring the denominator}} \cr & = \int {\frac{{3{x^2} + 12x - 20}}{{{{\left( {{x^2} - 4} \right)}^2}}}} dx \cr & = \int {\frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}}} dx \cr & {\text{The partial fraction decomposition of the integrand has the}} \cr & {\text{form}} \cr & \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{x + 2}} + \frac{D}{{{{\left( {x + 2} \right)}^2}}} \cr & {\text{Multiply both sides by the least common denominator}} \cr & 3{x^2} + 12x - 20 = A\left( {x - 2} \right){\left( {x + 2} \right)^2} + B{\left( {x + 2} \right)^2} \cr & + C\left( {x + 2} \right){\left( {x - 2} \right)^2} + D{\left( {x - 2} \right)^2} \cr & {\text{Expanding and simplifying we obtain}} \cr & 3{x^2} + 12x - 20 = A{x^3} + 2A{x^2} - 4Ax - 8A + B{x^2} + 4Bx + 4B \cr & + C{x^3} - 2C{x^2} - 4Cx + 8C + D{x^2} - 4Dx + 4D \cr & {\text{Collecting like terms}} \cr & 3{x^2} + 12x - 20 = \left( {A + C} \right){x^3} + \left( {2A + B - 2C + D} \right){x^2} \cr & + \left( { - 4A + 4B - 4C - 4D} \right)x + \left( { - 8A + 4B + 8C + 4D} \right) \cr & {\text{We obtain the following system of equations}} \cr & A + C = 0 \cr & 2A + B - 2C + D = 3 \cr & - 4A + 4B - 4C - 4D = 12 \cr & - 8A + 4B + 8C + 4D = - 20 \cr & {\text{Solving we get:}} \cr & A = 1,{\text{ }}B = 1,{\text{ }}C = - 1,{\text{ }}D = - 2 \cr & {\text{Therefore}}{\text{,}} \cr & \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{x + 2}} + \frac{D}{{{{\left( {x + 2} \right)}^2}}} \cr & \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{1}{{x - 2}} + \frac{1}{{{{\left( {x - 2} \right)}^2}}} - \frac{1}{{x + 2}} - \frac{2}{{{{\left( {x + 2} \right)}^2}}} \cr & = \int {\left[ {\frac{1}{{x - 2}} + \frac{1}{{{{\left( {x - 2} \right)}^2}}} - \frac{1}{{x + 2}} - \frac{2}{{{{\left( {x + 2} \right)}^2}}}} \right]} dx \cr & {\text{Integrating}} \cr & = \ln \left| {x - 2} \right| - \frac{1}{{x - 2}} - \ln \left| {x + 2} \right| + \frac{2}{{x + 2}} + C \cr & {\text{Simplifying}} \cr & = \ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{{x - 6}}{{{x^2} - 4}} + C \cr} $$
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