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: 47

Answer

\[ = \ln \left|\frac{{{\,{{\left( {x - 1} \right)}^2}} }}{{{x^2} + 4x + 5}}\right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}} dx \hfill \\ \hfill \\ Using\,\,partial\,fractions \hfill \\ \hfill \\ \frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4x + 5}} \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ \hfill \\ A = 2,{\text{ }}B = - 2,{\text{ C}} = - 10 \hfill \\ \hfill \\ \frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2x - 10}}{{{x^2} + 4x + 5}} \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2x + 4 - 14}}{{{x^2} + 4x + 4 + 1}} \hfill \\ \hfill \\ \frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right) - 14}}{{\,{{\left( {x + 2} \right)}^2} + 1}} \hfill \\ \hfill \\ \frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right)\,}}{{\,{{\left( {x + 2} \right)}^2} + 1}} - \frac{{14}}{{\,{{\left( {x + 2} \right)}^2} + 1}} \hfill \\ \hfill \\ substitute\,\,in\,\,\int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}} dx \hfill \\ \hfill \\ \int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}dx} \,\, = \int_{}^{} {\,\left( {\frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right)\,}}{{\,{{\left( {x + 2} \right)}^2} + 1}} - \frac{{14}}{{\,{{\left( {x + 2} \right)}^2} + 1}}} \right)dx} \hfill \\ \hfill \\ let\,\,u = \,{\left( {x + 2} \right)^2} + 1 \hfill \\ \hfill \\ = 2\ln \left| {x - 1} \right| - \int_{}^{} {\frac{{du}}{u}} + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) \hfill \\ \hfill \\ {\text{integrate}}\,\,\,{\text{and}}\,\,{\text{simplify}} \hfill \\ \hfill \\ = \ln \left| {\,{{\left( {x - 1} \right)}^2}} \right| - \ln \left| u \right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\ \hfill \\ = \ln \left| {\,{{\left( {x - 1} \right)}^2}} \right| - \ln \left| {\,{{\left( {x + 2} \right)}^2} + 1} \right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\ \hfill \\ = \ln \left|\frac{{{\,{{\left( {x - 1} \right)}^2}} }}{{{x^2} + 4x + 5}}\right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.