Chapter 10.7, Problem 47E

To determine

To calculate: Recognize partial fraction decomposition.

Answer to Problem 47E

Partial fraction decomposition of option (a) is given as below.

Explanation of Solution

Given information:

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

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

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

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

Calculation:

The factors of denominator of given partial fraction decomposition is as below,

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

In rational function,

  $r\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$

Where the degree of $P$ is less than the degree of $Q$ . Every polynomial with real coefficient can be factored completely into linear and irreducible quadratic factors.

The case where denominator is product of distinct linear factors,

  $Q\left(x\right)=\left(a_{1}x+b_1\right)\left(a_{2}x+b_2\right)\mathrm{...}\left(a_{n}x+b_n\right)$

The partial fraction decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$ takes the form,

  $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{a_{1}x+b_1}+A_2$

For remaining options (b), (C), (d)

The partial fraction decomposition $\frac{x}{{\left(x+1\right)}^2}$ is already a partial factor decomposition or whether it can be decomposed further.

The partial fraction decomposition $\frac{1}{x+1}+\frac{2}{{\left(x+1\right)}^2}$ is already partial factor decomposition or whether it can be decomposed further.

The partial fraction decomposition $\frac{x+2}{{\left(x^2+1\right)}^2}$ is already a partial factor decomposition or whether it can be decomposed further.