Q. 40 - Precalculus Enhanced with Graphing Utilities

Question

Write the partial fraction decomposition of each rational expression. 

$\frac{x^2+1}{x^3+x^2-5x+3}$

Step-by-Step Solution

Answer

The partial fraction decomposition is  

$\frac{x^2+1}{x^3+x^2-5x+3}=\frac{{\displaystyle \frac{3}{8}}}{x-1}+\frac{{\displaystyle \frac{1}{2}}}{{(x-1)}^2}+\frac{{\displaystyle \frac{5}{8}}}{x+3}$

Step 1. Given Information.

The rational expression is, 

$\frac{x^2+1}{x^3+x^2-5x+3}$

Step 2. Partial Fraction Decomposition.

Decompose the rational fraction, we get, 

$\frac{x^2+1}{x^3+x^2-5x+3}=\frac{x^2+1}{{(x-1)}^2(x+3)}$

$\frac{x^2+1}{{(x-1)}^2(x+3)}=\frac{A}{x-1}+\frac{B}{{(x-1)}^2}+\frac{C}{x+3}$

$\frac{x^2+1}{{(x-1)}^2(x+3)}=\frac{{(x-1)}^{2}C+(x-1)(x+3)A+(x+3)B}{{(x-1)}^2(x+3)}$

Therefore, 

$x^2+1=x^{2}A+x^{2}C+2xA+xB-2xC-3A+3B+C$

$x^2+1=x^2(A+C)+x(2A+B-2C)-3A+3B+C$

Equating the coefficients with the like powers to get, 

$$\begin{gathered} A+C=1 \\ 2A+B-2C=0 \\ -3A+3B+C=1 \end{gathered}$$

Step 3. Solving the equations.

we get,   

$$\begin{gathered} A=\frac{3}{8}, \\ B=\frac{1}{2}, \\ C=\frac{5}{8} \end{gathered}$$

Therefore,

$\frac{x^2+1}{x^3+x^2-5x+3}=\frac{{\displaystyle \frac{3}{8}}}{x-1}+\frac{{\displaystyle \frac{1}{2}}}{{(x-1)}^2}+\frac{{\displaystyle \frac{5}{8}}}{x+3}$