Answer
The partial fraction decomposition of the rational expression is
$\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$.
Work Step by Step
Factor the denominator by the method of grouping as the expression ${{x}^{2}}+x+1$ occurs multiple times in the denominator of the rational expression, so, for each power of ${{x}^{2}}+x+1$ assign an undefined constant given below:
$\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+x+1}+\frac{Cx+D}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$
Multiply both sides by ${{\left( {{x}^{2}}+x+1 \right)}^{2}}$ as given below:
$\begin{align}
& 4{{x}^{3}}+5{{x}^{2}}+7x-1=\left( Ax+B \right)\left( {{x}^{2}}+x+1 \right)+Cx+D \\
& 4{{x}^{3}}+5{{x}^{2}}+7x-1=A{{x}^{3}}+A{{x}^{2}}+Ax+B{{x}^{2}}+Bx+B+Cx+D \\
& 4{{x}^{3}}+5{{x}^{2}}+7x-1=A{{x}^{3}}+\left( A+B \right){{x}^{2}}+\left( A+B+C \right)x+\left( B+D \right) \\
\end{align}$
Then, equate the coefficients of like terms of the equation to write a system of equations as given below:
$ A=4$ (I)
$ A+B=5$ (II)
$ A+B+C=7$ (III)
$ B+D=-1$ (IV)
And from equation (II), obtain the value of B:
$\begin{align}
& B=5-4 \\
& =1
\end{align}$
From the equation (III), obtain the value of C:
$\begin{align}
& C=7-5 \\
& =2
\end{align}$
From the equation (IV), obtain the value of D:
$\begin{align}
& D=-1-1 \\
& =-2
\end{align}$
So, $ A=4,B=1,C=2,D=-2$. Put the values of $ A,\,\,B,\,\,C,\,\,\text{and }D $ in the given equation, and define the partial fraction decomposition:
$\frac{Ax+B}{{{x}^{2}}+x+1}+\frac{Cx+D}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$
Thus, the partial fraction decomposition of the rational expression is
$\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$.
