Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises: 25

Answer

\[ = \ln {\left| {\frac{{\,\left( {x + 1} \right)\,{{\left( {x - 3} \right)}^{\frac{1}{3}}}}}{{\,\left( {x - 1} \right)\,{{\left( {x + 3} \right)}^{\frac{1}{3}}}}}} \right|^{\frac{1}{{16}}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{{x^4} - 10{x^2} + 9}}} \hfill \\ \hfill \\ {\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ \frac{1}{{{x^4} - 10{x^2} + 9}} = \frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} \hfill \\ \hfill \\ Using\,\,partial\,fractions \hfill \\ \hfill \\ \frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x + 3}} + \frac{C}{{x - 1}} + \frac{D}{{x + 1}} \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ \hfill \\ A = - \frac{1}{{48}},\,\,B = \frac{1}{{48}},\,\,C = - \frac{1}{{16}},\,\,D = \frac{1}{{16}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} = \frac{{ - \frac{1}{{48}}}}{{x + 3}} + \frac{{\frac{1}{{48}}}}{{x - 3}} + \frac{{ - \frac{1}{{16}}}}{{x - 1}} + \frac{{\frac{1}{{16}}}}{{x + 1}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{{x^4} - 10{x^2} + 9}}} = \int_{}^{} {\,\left( {\frac{{ - \frac{1}{{16}}}}{{x - 1}} + \frac{{\frac{1}{{16}}}}{{x + 1}} + \frac{{ - \frac{1}{{48}}}}{{x + 3}} + \frac{{\frac{1}{{48}}}}{{x - 3}}} \right)} \hfill \\ \hfill \\ {\text{integrating}} \hfill \\ \hfill \\ = - \frac{1}{{16}}\ln \left| {x - 1} \right| + \frac{1}{{16}}\ln \left| {x + 1} \right| - \frac{1}{{48}}\ln \left| {x + 3} \right| + \frac{1}{{48}}\ln \left| {x - 3} \right| + C \hfill \\ \hfill \\ = \ln {\left| {\frac{{\,\left( {x + 1} \right)\,{{\left( {x - 3} \right)}^{\frac{1}{3}}}}}{{\,\left( {x - 1} \right)\,{{\left( {x + 3} \right)}^{\frac{1}{3}}}}}} \right|^{\frac{1}{{16}}}} + C \hfill \\ \hfill \\ \end{gathered} \]
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