Chapter 10.7, Problem 41E

To determine

To calculate: Partial fraction decomposition of the rational function.

Answer to Problem 41E

Partial fraction decomposition of the rational function is $\frac{1}{x}+\frac{1}{x^2+1}+\frac{-x-2}{{\left(x^2+1\right)}^2}$ .

Explanation of Solution

Given information:

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

Calculation:

It is given that,

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

The form of the partial fraction decomposition is,

  $\frac{x^4+x^3+x^2-x+1}{x{\left(x^2+1\right)}^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{{\left(x^2+1\right)}^2}$

Right hand side as a single fraction,

  $\frac{x^4+x^3+x^2-x+1}{x{\left(x^2+1\right)}^2}=\frac{x\left(x^2+1\right)\left(Bx+C\right)+x\left(Dx+E\right)+{\left(x^2+1\right)}^{2}A}{x{\left(x^2+1\right)}^2}$

Denominator are equal, so one need equality of the numerator:

  $x^4+x^3+x^2-x+1=x\left(x^2+1\right)\left(Bx+C\right)+x\left(Dx+E\right)+{\left(x^2+1\right)}^{2}A$

Now just expand right hand side

  $x^4+x^3+x^2-x+1=x^{4}A+x^{3}B+x^{3}C+2x^{2}A+x^{2}B+x^{2}D+xC+xE+A$

Collect up all the like terms,

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

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

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

After solving this one can get the,

  $A=1,B=0,C=1,D=-1,E=-2$

So,

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