Chapter 7.4, Problem 40E

To determine

To find: write the partial fraction of the given rational expression $\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}$

Answer to Problem 40E

The final answer is $\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}=\frac{1}{{\left(x^2+4\right)}^2}+\frac{-4x+2}{{\left(x^2+4\right)}^3}$

Explanation of Solution

Given information: we have the given rational fraction is $\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}$

Let the partial fraction form of the rational expression is

  $\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}=\frac{Ax+B}{\left(x^2+4\right)}+\frac{Cx+D}{{\left(x^2+4\right)}^2}+\frac{Ex+F}{{\left(x^2+4\right)}^3}$

Now we solve the equation for the constants A,B,C, D,E and F

Now, the above equation is equivalent to

  $\left(Ax+B\right){\left(x^2+4\right)}^2+\left(Cx+D\right)\left(x^2+4\right)+\left(Ex+F\right)=x^2-4x+6$

Expanding the L.H.S we get

  $A{x}^5+8A{x}^3+16Ax+B{x}^4+8B{x}^2+16B+C{x}^3+4Cx+D{x}^2+4D+Ex+F=x^2-4x+6$

Grouping element’s according powers of x we get

  $A{x}^5+B{x}^4+\left(8A+C\right)x^3+\left(8B+D\right)x^2+\left(16A+4C+E\right)x+\left(16B+4D+F\right)=x^2-4x+6$

Now, equating the coefficients of similar term on both sides to make a list of equations

  $\begin{array}{c}16B+4D+F=6\\ 16A+4C+E=-4\\ 8B+D=1\\ 8A+C=0\\ A=0\\ B=0\end{array}$

Solving the above system we get

  $A=0,B=0,C=0,D=1,E=-4andF=2$

Therefore, the required partial fraction is

  $\begin{array}{c}\frac{x^2-4x+6}{{\left(x^2+4\right)}^3}=\frac{0x+0}{\left(x^2+4\right)}+\frac{0x+1}{{\left(x^2+4\right)}^2}+\frac{-4x+2}{{\left(x^2+4\right)}^3}\\ =\frac{1}{{\left(x^2+4\right)}^2}+\frac{-4x+2}{{\left(x^2+4\right)}^3}\end{array}$