Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 15

Answer

$$4\ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + 4{\tan ^{ - 1}}x + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{3{x^2} + 4x + 4}}{{{x^3} + x}}} dx \cr & {\text{Decompose the integrand into partial fractions.}} \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{{3{x^2} + 4x + 4}}{{{x^3} + x}} = \frac{{3{x^2} + 4x + 4}}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr & \cr & {\text{Multiplying by }}x\left( {{x^2} + 1} \right){\text{, we have}} \cr & 3{x^2} + 4x + 4 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x \cr & 3{x^2} + 4x + 4 = A{x^2} + A + B{x^2} + Cx \cr & \cr & {\text{Group like-terms}} \cr & 3{x^2} + 4x + 4 = \left( {A{x^2} + B{x^2}} \right) + Cx + A \cr & {\text{comparing coefficients}}{\text{, we get the system}} \cr & A + B = 3,\,\,\,\,C = 4,\,\,A = 4 \cr & \cr & {\text{Solving, we obtain}} \cr & A = 4,\,\,\,B = - 1,\,\,\,C = 4 \cr & \cr & {\text{The decomposition of the integrand is}} \cr & \frac{{3{x^2} + 4x + 4}}{{x\left( {{x^2} + 1} \right)}} = \frac{4}{x} + \frac{{ - x + 4}}{{{x^2} + 1}} \cr & \frac{{3{x^2} + 4x + 4}}{{x\left( {{x^2} + 1} \right)}} = \frac{4}{x} - \frac{x}{{{x^2} + 1}} + \frac{4}{{{x^2} + 1}} \cr & \cr & \int {\frac{{3{x^2} + 4x + 4}}{{{x^3} + x}}} dx = \int {\frac{4}{x}dx - \int {\frac{x}{{{x^2} + 1}}dx} + \int {\frac{4}{{{x^2} + 1}}} dx} \cr & {\text{Integrating}} \cr & = 4\ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + 4{\tan ^{ - 1}}x + C \cr} $$
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