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: 64

Answer

\[ = \frac{{{x^2}}}{2} + \frac{1}{9}\ln \left| x \right| - \frac{{41}}{9}\ln \left( {{x^2} + 9} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{x^4} + 1}}{{{x^3} + 9x}}\,} dx \hfill \\ \hfill \\ Given\,\,function \hfill \\ \hfill \\ \frac{{{x^4} + 1}}{{{x^3} + 9x}} \hfill \\ \hfill \\ performing\,\,long\,\,division \hfill \\ \hfill \\ \frac{{{x^4} + 1}}{{{x^3} + 9x}} = x + \frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} \hfill \\ \hfill \\ Using\,\,partial\,\,fraction\,\,for\,\,\,\frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} \hfill \\ \hfill \\ \frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} = \frac{A}{{9x}} + \frac{{Bx + C}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ A = 1,{\text{ B}} = - 82/9{\text{ and }}B = 0 \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} = \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\ \hfill \\ \frac{{{x^4} + 1}}{{{x^3} + 9x}} = x + \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{{x^4} + 1}}{{{x^3} + 9x}}\,} dx = \int_{}^{} {\,\left( {x + \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}}} \right)\,dx} \hfill \\ \hfill \\ \textbf{ integrate} \hfill \\ \hfill \\ =\boxed{ \frac{{{x^2}}}{2} + \frac{1}{9}\ln \left| x \right| - \frac{{41}}{9}\ln \left( {{x^2} + 9} \right) + C} \hfill \\ \end{gathered} \]
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