Chapter 7.4, Problem 29E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 29E

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

Explanation of Solution

Given information: $\frac{x^2+2x+3}{x^3+x}$

Concept Involved:

Decomposition of $N\left(x\right)/D\left(x\right)$ into Partial Fractions:

1. Divide when improper: When $N\left(x\right)/D\left(x\right)$ is an improper fraction [degree ofN(x) = degree of D(x)], divide the denominator into the numerator to obtain

$\frac{N\left(x\right)}{D\left(x\right)}=\left(polynomial\right)+\frac{N_1\left(x\right)}{D\left(x\right)}$

and apply Steps 2, 3, and 4 to the proper rational expression $\frac{N_1\left(x\right)}{D\left(x\right)}$

Note that $N_1\left(x\right)$ is the remainder from the division of $N\left(x\right)$ by $D\left(x\right)$ .

2. Factor the denominator: Completely factor the denominator into factors of the form ${\left(px+q\right)}^m$ and ${\left(a{x}^2+bx+c\right)}^n$ where $\left(a{x}^2+bx+c\right)$ is irreducible
3. Linear factors: For each factor of the form ${\left(px+q\right)}^m$ , the partial fraction decomposition must include the following sum of $m$ fractions.

$\frac{A_1}{\left(px+q\right)}+\frac{A_2}{{\left(px+q\right)}^2}+\mathrm{.....}+\frac{A_m}{{\left(px+q\right)}^m}$

4. Quadratic factors: For each factor of the form ${\left(a{x}^2+bx+c\right)}^n$, the partial fraction decomposition must include the following sum of n fractions.

$\frac{B_{1}x+C_1}{a{x}^2+bx+c}+\frac{B_{2}x+C_2}{{\left(a{x}^2+bx+c\right)}^2}+\mathrm{.....}+\frac{B_{n}x+C_n}{{\left(a{x}^2+bx+c\right)}^n}$

Calculation:

Factor the GCF of the denominator

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

Write one partial fraction for each of those factors

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

Multiply through by the LCD $x\left(x^2+1\right)$ so we no longer have fractions

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

Cancel common factors throughout the equation

  $x^2+2x+3=A\left(x^2+1\right)+\left(Bx+C\right)x$

Distribute in right side of the equation

  $x^2+2x+3=A{x}^2+A+B{x}^2+Cx$

Group $x^2,x,\&$ constant terms in right side of the equation

  $x^2+2x+3=A{x}^2+B{x}^2+Cx+A$

Factor $x^2,x,\&$ constant terms in right side of the equation

  $1x^2+2x+3=\left(A+B\right)x^2+Cx+A$

Equate the coefficient of $x^2,x,\&$ constant in both sides of the equation

  $A+B=1$

  $C=2$

  $A=3$

Label the equation we got

  $A+B=1$  ▶ 1^st equation

  $C=2$  ▶ 2^nd equation

  $A=3$  ▶ 3^rd equation

Substitute 3 for A in 1^st equation and solve for B

• By subtracting 3 on both sides, then
• Combining like terms in both sides of the equation

  $3+B=1$

  $3+B-3=1-3$

  $B=-2$

Substitute the value of A, B, and C in the equation $\frac{x^2+2x+3}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$

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

Conclusion:

The partial fraction decomposition of the given rational expressionis: $\boxed{\frac{3}{x}-\frac{2x}{x^2+1}+\frac{2}{x^2+1}}$