Chapter 11, Problem 36RE

To determine

To find: The partial fraction decomposition .

Answer to Problem 36RE

  $\frac{x}{(x^2+9)(x+1)}=\frac{\frac{-1}{10}}{(x+1)}+\frac{\frac{x}{10}+\frac{9}{10}}{(x^2+9)}$

Explanation of Solution

Given information: $\frac{x}{(x^2+9)(x+1)}$

  $\begin{array}{l}\frac{x}{(x^2+9)(x+1)}=\frac{A}{(x+1)}+\frac{Bx+c}{x^2+9}\\ \Rightarrow \frac{x}{(x^2+9)(x+1)}=\frac{A(x^2+9)+(Bx+c)(x+1)}{(x^2+9)(x+1)}\\ \Rightarrow x=A(x^2+9)+(Bx+c)(x+1)\end{array}$

Put $x=-1$

  $A=\frac{-1}{10}$

Now from equation (1)

  $\begin{array}{l}\Rightarrow x=A(x^2+9)+(Bx+c)(x+1)\\ \Rightarrow x=\frac{-1}{10}(x^2+9)+B{x}^2+Bx+cx+c\\ \Rightarrow x=\frac{-x^2}{10}-\frac{9}{10}+B{x}^2+Bx+cx+c\\ \Rightarrow x=x^2(B-\frac{1}{10})+x(B+C)+c-\frac{9}{10}\end{array}$

The L.H.S and R.H.S will be equal when

  $\begin{array}{l}B-\frac{1}{10}=0\\ \therefore B=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\end{array}$

And

  $\begin{array}{l}c-\frac{9}{10}=0\\ \therefore c=\frac{9}{10}\end{array}$

And $B+C=1$

  $\therefore$ partial fraction ,

  $\frac{x}{(x^2+9)(x+1)}=\frac{A}{(x+1)}+\frac{Bx+c}{x^2+9}$

  $\Rightarrow$ $\frac{x}{(x^2+9)(x+1)}=\frac{\frac{-1}{10}}{(x+1)}+\frac{\frac{x}{10}+\frac{9}{10}}{(x^2+9)}$