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: 20

Answer

\[ = - \frac{1}{{18}}\ln \left| y \right| - \frac{5}{{54}}\ln \left| {y + 6} \right| + \frac{4}{{27}}\ln \left| {y - 3} \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{y + 1}}{{{y^3} + 3{y^2} - 18y}}} \hfill \\ \hfill \\ factor\,the\,denominator. \hfill \\ \hfill \\ \frac{{y + 1}}{{y\,\left( {{y^2} + 3y - 18} \right)}} = \frac{{y + 1}}{{y\,\left( {y + 6} \right)\,\left( {y - 3} \right)}} \hfill \\ \hfill \\ decomposition\,\,into\,\,partial\,\,fraction. \hfill \\ \hfill \\ \frac{{y + 1}}{{y\,\left( {y + 6} \right)\,\left( {y - 3} \right)}} = \frac{A}{y} + \frac{B}{{y + 6}} + \frac{C}{{y - 3}} \hfill \\ \hfill \\ simplify\,\,the\,\,{\text{denominators}} \hfill \\ \hfill \\ y + 1 = A\,\left( {y + 6} \right)\,\left( {y - 3} \right) + By\,\left( {y - 3} \right) + Cy\,\left( {y + 6} \right) \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ \hfill \\ y = 0\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,A = - \frac{1}{{18}} \hfill \\ y = - 6\,\,\,\, \to \,\,\,\,\,B = - \frac{5}{{54}} \hfill \\ y = 3\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,C = \frac{4}{{27}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{{ - \frac{1}{{18}}}}{y}dy + \int_{}^{} {\frac{{ - \frac{5}{{54}}}}{{y + 6}}dy + \int_{}^{} {\frac{{\frac{4}{{27}}}}{{y - 3}}dy} } } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = - \frac{1}{{18}}\ln \left| y \right| - \frac{5}{{54}}\ln \left| {y + 6} \right| + \frac{4}{{27}}\ln \left| {y - 3} \right| + C \hfill \\ \end{gathered} \]
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