## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 10

#### Answer

The partial fraction function is $-\frac{1}{x}+\frac{1}{\left( x-1 \right)}$.

#### Work Step by Step

We find the partial fraction of the provided expression as given below, $\frac{1}{x\left( x-1 \right)}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}$ Multiply $x\left( x-1 \right)$ on both sides, \begin{align} & x\left( x-1 \right)\frac{1}{x\left( x-1 \right)}=x\left( x-1 \right)\frac{A}{x}+x\left( x-1 \right)\frac{B}{\left( x-1 \right)} \\ & 1=\left( x-1 \right)A+xB \\ & =Ax-A+Bx \\ & 1=x\left( A+B \right)-A \end{align} …… (I) Now, compare the coefficient of equation (I), $A+B=0$ …… (II) \begin{align} & -A=1 \\ & A=-1 \\ \end{align} …… (III) Now, substitute the value of equation (III) into equation (II), \begin{align} & -1+B=0 \\ & B=1 \end{align} Therefore, $\frac{1}{x\left( x-1 \right)}=\frac{-1}{x}+\frac{1}{\left( x-1 \right)}$ Thus, the partial fraction of the provided expression is $\frac{-1}{x}+\frac{1}{\left( x-1 \right)}$.

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