Chapter 7.3, Problem 67E

To determine

To write the Partial fraction decomposition of the rational expression $\frac{4x^2+2x-1}{x^2\left(x+1\right)}$

Answer to Problem 67E

The partial fraction of expansion is $\frac{3x-1}{x^2}+\frac{1}{x+1}$

Explanation of Solution

Given information:

The rational expression is $\frac{4x^2+2x-1}{x^2\left(x+1\right)}$

The rational expression is $\frac{4x^2+2x-1}{x^2\left(x+1\right)}$

Here the denominator contain linear and quadratic term as factors and the given rational expression can be expressed as sum of partial fraction.

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

where $A$ , $B$ and $C$ are the coefficients.

Now , for finding the value of coefficients $A$ , $B$ and $C$

  $\begin{array}{l}\frac{4x^2+2x-1}{x^2\left(x+1\right)}=\frac{Ax+B}{x^2}+\frac{C}{x+1}=\frac{(Ax+B)\left(x+1\right)+x^{2}C}{x^2\left(x+1\right)}=\frac{x^2\left(A+C\right)+x\left(A+B\right)+B}{x^2\left(x+1\right)}\\ 4x^2+2x-1=x^2\left(A+C\right)+x\left(A+B\right)+B\end{array}$

Comparing constant terms in the equation

  $B=-1$

Comparing coefficients of $x$

  $\begin{array}{l}A+B=2\\ A=2-B\\ A=3\end{array}$

Comparing coefficients of $x^2$

  $\begin{array}{l}A+C=4\\ C=4-A\\ C=1\end{array}$

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

Checking algebraically,

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

Checking Graphically,

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

  

  $\frac{3x-1}{x^2}+\frac{1}{x+1}$

  

The graph of rational expression and partial fraction is same hence graphically it is proved that $\frac{4x^2+2x-1}{x^2\left(x+1\right)}=\frac{3x-1}{x^2}+\frac{1}{x+1}$