Chapter 7, Problem 58RE

To determine

To calculate: The partial fraction of the rational expression.

Answer to Problem 58RE

The partial fraction is $\frac{279}{260\left(x+16\right)}+\frac{-19x+44}{260\left(x^2+4\right)}$ .

Explanation of Solution

Given information:

The given rational expression is $\frac{x^2-x+7}{x^4+8x^2+16}$ .

Calculation:

Consider the given rational expression is $\frac{x^2-x+7}{x^4+8x^2+16}$ .

  $\begin{array}{c}\frac{x^2-x+7}{x^4+8x^2+16}=\frac{x^2-x+7}{x^4+4x^2+4x^2+16}\\ =\frac{x^2-x+7}{x^2\left(x^2+4\right)+4\left(x+16\right)}\\ =\frac{x^2-x+7}{\left(x+16\right)\left(x^2+4\right)}\\ =\frac{A}{\left(x+16\right)}+\frac{Bx+C}{\left(x^2+4\right)}\end{array}$  ...... (1)

Now, find $A,BandC$ .

  $\begin{array}{c}\frac{x^2-x+7}{x^4+8x^2+16}=\frac{A}{\left(x+16\right)}+\frac{Bx+C}{\left(x^2+4\right)}\\ \frac{x^2-x+7}{x^4+8x^2+16}=\frac{A\left(x^2+4\right)+Bx\left(x+16\right)+C\left(x+16\right)}{\left(x+16\right)\left(x^2+4\right)}\\ \frac{x^2-x+7}{x^4+8x^2+16}=\frac{A{x}^2+4A+B{x}^2+16Bx+Cx+16C}{x^4+8x^2+16}\\ x^2-x+7=A{x}^2+4A+B{x}^2+16Bx+Cx+16C\\ x^2-x+7=x^2\left(A+B\right)+x\left(16B+C\right)+4A+16C\end{array}$

Now, equate the coefficients.

  $A+B=1$  ...... (2)

  $16B+C=-1$  ...... (3)

  $4A+16C=7$  ...... (4)

Now, multiply 4 with the equation (2) and subtract from equation (4).

  $\begin{array}{c}4A+16C-4A-4B=7-4\\ 16C-4B=3\end{array}$  ...... (6)

Now, multiply 4 in the equation (6) and add it to the equation (3).

  $\begin{array}{c}16B+C+64C-16B=-1+12\\ 65C=11\\ C=\frac{11}{65}\end{array}$

Substitute the value $C=\frac{11}{65}$ in the equation (4).

  $\begin{array}{c}4A+16\times \frac{11}{65}=7\\ 4A+\frac{176}{65}=7\\ A=\frac{279}{260}\end{array}$

Similarly, substitute the value $C=\frac{11}{65}$ in the equation (3).

  $\begin{array}{c}16B+\frac{11}{65}=-1\\ 16B=-\frac{76}{65}\\ B=-\frac{19}{260}\end{array}$

Now, substitute the value of $A,BandC$ in the equation (1).

  $\begin{array}{c}\frac{x^2-x+7}{x^4+8x^2+16}=\frac{\frac{279}{260}}{\left(x+16\right)}+\frac{-\frac{19}{260}x+\frac{11}{65}}{\left(x^2+4\right)}\\ =\frac{279}{260\left(x+16\right)}+\frac{-19x+44}{260\left(x^2+4\right)}\end{array}$

Now, draw the graph.

  

Hence, the required partial fraction is $\frac{279}{260\left(x+16\right)}+\frac{-19x+44}{260\left(x^2+4\right)}$ .