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: 22


The partial fraction is, $\frac{1}{\left( x+1 \right)}-\frac{1}{{{\left( x+1 \right)}^{2}}}$

Work Step by Step

$\frac{x}{{{\left( x+1 \right)}^{2}}}=\frac{A}{\left( x+1 \right)}+\frac{B}{{{\left( x+1 \right)}^{2}}}$ Now, multiply both sides by ${{\left( x+1 \right)}^{2}}$: ${{\left( x+1 \right)}^{2}}\times \frac{x}{{{\left( x+1 \right)}^{2}}}={{\left( x+1 \right)}^{2}}\times \frac{A}{\left( x+1 \right)}+{{\left( x+1 \right)}^{2}}\times \frac{B}{{{\left( x+1 \right)}^{2}}}$ $\begin{align} & x=A\left( x+1 \right)+B \\ & =Ax+A+B \end{align}$ Then, compare the coefficient of $x$ and constant term: $A=1$ …… (1) $A+B=0$ …… (2) Then, put the value of A in equation (2): $\begin{align} & 1+B=0 \\ & B=-1 \end{align}$ Therefore, $\frac{x}{{{\left( x+1 \right)}^{2}}}=\frac{1}{\left( x+1 \right)}-\frac{1}{{{\left( x+1 \right)}^{2}}}$ Thus, the partial fraction of the given expression is $\frac{1}{\left( x+1 \right)}-\frac{1}{{{\left( x+1 \right)}^{2}}}$.
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