Question

Write the partial fraction decomposition of each rational expression.

Short answer

The partial fraction decomposition is,

$\frac{x^3+1}{x^4(x-1)}=\frac{-2}{x}+\frac{-1}{x^2}+\frac{-1}{x^3}+\frac{-1}{x^4}+\frac{2}{x-1}$

Step-by-step solution

Step 1. Given Information. 

The rational expression is,

$\frac{x^3+1}{x^5-x^4}$

Step 2. Partial Fraction Decomposition. 

Decompose the rational fraction, we get,

$\frac{x^3+1}{x^4(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x-1}$

$\frac{x^3+1}{x^4(x-1)}=\frac{x^{4}E+x^3(x-1)A+x^2(x-1)B+x(x-1)C+(x-1)D}{x^4(x-1)}$

Therefore,

$x^3+1=x^{4}E+x^3(x-1)A+x^2(x-1)B+x(x-1)C+(x-1)D$

$x^3+1=x^4(A+E)+x^3(-A+B)+x^2(-B+C)+x(-C+D)-D$

Equating the coefficients with the like powers to get,

$$\begin{gathered} A+E=0 \\ -A+B=1 \\ -B+C=0 \\ -C+D=0 \\ -D=1 \end{gathered}$$

Step 3. Solving the equations. 

we get,

$A=-2,B=-1,C=-1,D=-1,E=2$

Therefore,

$\frac{x^3+1}{x^4(x-1)}=\frac{-2}{x}+\frac{-1}{x^2}+\frac{-1}{x^3}+\frac{-1}{x^4}+\frac{2}{x-1}$

$A=-2,B=-1,C=-1,D=-1,E=2$