Answer
The partial fraction decomposition of the expression $\frac{1}{x\left( {{x}^{2}}+x+1 \right)}$ is $\frac{1}{x}-\frac{x+1}{{{x}^{2}}+x+1}$.
Work Step by Step
Considered the expression, $\frac{1}{x\left( {{x}^{2}}+x+1 \right)}$
Step 1. Set up the partial fraction decomposition with the unknown constants.
Now, use the partial fraction decomposition property for linear and quadratic factors:
$\frac{1}{x\left( {{x}^{2}}+x+1 \right)}=\frac{A}{x}+\frac{Bx+C}{\left( {{x}^{2}}+x+1 \right)}$
Step 2. Multiply both sides of the resulting equation by the least common denominator.
Now, multiply both sides by $x\left( {{x}^{2}}+x+1 \right)$:
$\begin{align}
& x\left( {{x}^{2}}+x+1 \right)\left( \frac{1}{x\left( {{x}^{2}}+x+1 \right)} \right)=x\left( {{x}^{2}}+x+1 \right)\left( \frac{A}{x}+\frac{Bx+C}{\left( {{x}^{2}}+x+1 \right)} \right) \\
& x\left( {{x}^{2}}+x+1 \right)\cdot \frac{1}{x\left( {{x}^{2}}+x+1 \right)}=x\left( {{x}^{2}}+x+1 \right)\cdot \frac{A}{x}+x\left( {{x}^{2}}+x+1 \right)\cdot \frac{Bx+C}{\left( {{x}^{2}}+x+1 \right)} \\
& 1=A\left( {{x}^{2}}+x+1 \right)+\left( Bx+C \right)x
\end{align}$
Step 3. Simplify the right side of the equation.
$1=A{{x}^{2}}+Ax+A+B{{x}^{2}}+Cx$
Step 4. Write both sides by descending powers, equate coefficients of like powers of $x$, and equate the constant term.
$1=\left( A+B \right){{x}^{2}}+\left( A+C \right)x+A$
Now, equate the constant term:
$A+B=0$, $A+C=0$ and $A=1$
Step 5. Solve the resulting linear system.
Put the value of $A=1$ in $A+B=0$ and $A+C=0$ ,
$\begin{align}
& 1+B=0 \\
& B=-1
\end{align}$
And
$\begin{align}
& 1+C=0 \\
& C=-1
\end{align}$
Thus, the values of the constants are $A=1,\text{ }B=-1\text{ and }C=-1$.
Step 6. Substitute the values of A, B and C, and write the partial fraction decomposition.
Now,
$\begin{align}
& \frac{1}{x\left( {{x}^{2}}+x+1 \right)}=\frac{\left( 1 \right)}{x}+\frac{\left( -1 \right)x+\left( -1 \right)}{\left( {{x}^{2}}+x+1 \right)} \\
& =\frac{1}{x}+\frac{-x-1}{\left( {{x}^{2}}+x+1 \right)} \\
& =\frac{1}{x}-\frac{x+1}{\left( {{x}^{2}}+x+1 \right)}
\end{align}$
Hence, the partial fraction decomposition of the expression $\frac{1}{x\left( {{x}^{2}}+x+1 \right)}$ is $\frac{1}{x}-\frac{x+1}{{{x}^{2}}+x+1}$.
