Chapter 7.3, Problem 61E

To determine

To write the Partial fraction decomposition of the rational expression $\frac{1}{x^2+x}$

Answer to Problem 61E

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

Explanation of Solution

Given information:

The rational expression is $\frac{1}{x^2+x}$

The rational expression is $\frac{1}{x^2+x}$

Here the denominator can be expressed in terms of product of linear factors and the given rational expression can be expressed as sum of partial fraction.

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

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

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

  $\begin{array}{l}1=A\left(x+1\right)+Bx\\ 1=x\left(A+B\right)+A\end{array}$   s

Comparing constant terms in the equation

  $A=1$

Comparing coefficients of $x$

  $\begin{array}{l}A+B=0\\ B=-A\\ B=-1\end{array}$

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

Checking algebraically,

  $\frac{1}{x}+\frac{-1}{x+1}=\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}$

Checking Graphically,

  $\frac{1}{x^2+x}$

  

  $\frac{1}{x}+\frac{-1}{x+1}$

  

Interpretation:

The graph of rational expression and partial fraction is same hence graphically it is proved that

  $\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{-1}{x+1}$