Chapter 10.7, Problem 27E

To determine

The partial decomposition of the given rational function is to be estimated:

  $\frac{x^2+1}{x^3+x^2}$

Answer to Problem 27E

The partial decomposition of $\frac{x^2+1}{x^3+x^2}$ is to be estimated as: $\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}$

Explanation of Solution

Given:

  $\frac{x^2+1}{x^3+x^2}$

Concept Used:

The process of starting the given rational number with simplified answer steps is the partial-fraction decomposition.

Calculation:

First, need to simply the given expression:

  $\frac{x^2+1}{x^3+x^2}=\frac{x^2+1}{x^2\left(x+1\right)}$

The form of the partial decomposition is expressed by the following expression:

  $\frac{x^2+1}{\left(x^2\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$

Now, write the right hand side of the equation into a single fraction:

  $\frac{x^2+1}{x\left(x^2\right)\left(x+1\right)}=\frac{x^{2}C+x\left(x+1\right)A+\left(x+1\right)B}{\left(x^2\right)\left(x+1\right)}$

As the denominators of both eqns. are equal so use the equality of numerators:

  $\begin{array}{l}{x}^2+1=x^{2}C+x\left(x+1\right)A+\left(x+1\right)B\\ x^2+1=x^{2}C+x^{2}A+xA+xB+B\\ x^2+1=x^2\left(A+C\right)+x\left(A+B\right)+B\end{array}$

The coefficients near the like terms should be equal, so the following system is obtained:

  $\begin{array}{l}A+C=1\mathrm{...}\left(1\right)\\ A+B=0\mathrm{...}(2)\\ B=1\mathrm{...}\left(3\right)\end{array}$

The value of $A$ :

  $\begin{array}{l}A+B=0\\ A+1=0\\ A=-1\end{array}$

The value of $C$ :

  $\begin{array}{l}A+C=1\\ -1+C=1\\ C=2\end{array}$

Therefore, the partial decomposition of the given rational function is to be estimated as:

  $\begin{array}{l}\frac{x^2+1}{\left(x^2\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\\ \frac{x^2+1}{\left(x^2\right)\left(x+1\right)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}\\ \frac{x^2+1}{x^3+x^2}=\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}\end{array}$

Conclusion:

Hence, $\frac{x^2+1}{x^3+x^2}=\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}$