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: 38

Answer

\[ = \frac{A}{x} + \frac{B}{{\,\left( {x - 3} \right)}} + \frac{C}{{\,{{\left( {x - 3} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} \frac{2}{{x\,\left( {{x^2} - 6x + 9} \right)}} \hfill \\ \hfill \\ {\text{factor}}\,\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ \frac{2}{{x\,\left( {{x^2} + 6x + 9} \right)}} = \frac{2}{{x\,\left( {{x^2} - 2\,\left( 3 \right)\,\left( x \right) + {3^2}} \right)}} = \frac{2}{{x\,{{\left( {x - 3} \right)}^2}}} \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ we\,can\,\,use\,partial\,fraction\,of\,the\,form \hfill \\ \hfill \\ \frac{2}{{x\,{{\left( {x - 3} \right)}^2}}} = \frac{A}{x} + \frac{B}{{\,\left( {x - 3} \right)}} + \frac{C}{{\,{{\left( {x - 3} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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