Chapter 11, Problem 38RE

To determine

To find: The partial fraction decomposition .

Answer to Problem 38RE

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

Explanation of Solution

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

  $\begin{array}{l}\frac{x^2}{(x^2+)(x+1)(x-1)}=\frac{A}{(x^2+1)}+\frac{B}{x+1}+\frac{C}{x-1}\mathrm{......}\text{(1)}\\ \Rightarrow \frac{x^2}{(x^2+)(x+1)(x-1)}=\frac{A(x+1)(x-1)+B(x^2+1)(x-1)+C(x^2+1)(x+1)}{(x^2+)(x+1)(x-1)}\\ \Rightarrow x^2=A(x+1)(x-1)+B(x^2+1)(x-1)+C(x^2+1)(x+1)\mathrm{......}\text{(1)}\end{array}$

Put $x=-1$ in equation

  $\begin{array}{l}$ {(-1)}^2=AXO+B[{(-1)}^2+1](-1-1)+C\times 0$\Rightarrow 1=B(2)(-2)\\ \therefore B=\frac{-1}{4}\end{array}$

Now putting the value of B and C in equation (2), we get ,

  $\begin{array}{l}\Rightarrow x^2=A(x^2-1)-\frac{1}{4}(x^2+1)(x-1)+\frac{1}{4}(x^2+1)(x+1)\\ \Rightarrow x^2=A{x}^2-A-\frac{1}{4}(x^3-x^2-x-1)+\frac{1}{4}(x^3+x^2+x+1)\\ \Rightarrow x^2=A{x}^2-A-\frac{x^3}{4}+\frac{x^2}{4}-\frac{x}{4}+\frac{1}{4}+\frac{x^3}{4}+\frac{x^2}{4}+\frac{x}{4}+\frac{1}{4}\\ \Rightarrow x^2=A{x}^2+\frac{2x^2}{4}-A+\frac{1}{4}+\frac{1}{4}\\ \Rightarrow x^2=x^2(A+\frac{1}{2})-A+\frac{1}{2}\end{array}$

L.H.S=R.H.S

When

  $A+\frac{1}{2}=1$

And $-A+\frac{1}{2}=0$

  $\therefore$ Required $A=\frac{1}{2}$

Now putting the value of A, B and C in equation (1)

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

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

  $A=\frac{-1}{10}$

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 ,

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

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