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: 41

Answer

\[ = \frac{A}{{x - 4}} + \frac{B}{{\,{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3x + 4}}\]

Work Step by Step

\[\begin{gathered} \frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\ \hfill \\ {\text{factor}}\,\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ \frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} = \frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 2\,\left( 4 \right)x + {4^2}} \right)\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\ \hfill \\ = \frac{{2{x^2} + 3}}{{\,{{\left( {x - 4} \right)}^2}\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\ \hfill \\ we\,can\,\,use\,partial\,fraction\,of\,the\,form \hfill \\ \hfill \\ \frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{\,{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3x + 4}} \hfill \\ \end{gathered} \]
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