Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 33

Answer

$ \frac{1}{4x}+\frac{1}{x^2}-\frac{x+4}{4(x^2+4)}$

Work Step by Step

Step 1. Assume the rational function can be decomposed into $\frac{x+4}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$ Step 2. Evaluate the right side; we have $RHS=\frac{Ax(x^2+4)+B(x^2+4)+x^2(Cx+D)}{x^2(x^2+4)}==\frac{(A+C)x^3+(B+D)x^2+(4A)x+4B}{x^2(x^2+4)}$ Step 3. Comparing with the LHS, we can set up the following: $\begin{cases} A+C=0\\ B+D=0 \\ 4A=1 \\4B=4 \end{cases}$ Step 4. We can find the solutions as $A=\frac{1}{4}, B=1, C=-\frac{1}{4}, D=-1$ Step 5. The decomposition results is: $\frac{x+4}{x^2(x^2+4)}=\frac{1}{4x}+\frac{1}{x^2}-\frac{x+4}{4(x^2+4)}$
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