Answer
The partial fraction is, $\frac{4}{\left( x+1 \right)}+\frac{2x-3}{\left( {{x}^{2}}+1 \right)}$.
Work Step by Step
$\begin{align}
& \frac{6{{x}^{2}}-x+1}{{{x}^{3}}+{{x}^{2}}+x+1}=\frac{6{{x}^{2}}-x+1}{{{x}^{2}}\left( x+1 \right)+\left( x+1 \right)} \\
& =\frac{6{{x}^{2}}-x+1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)} \\
& =\frac{A}{\left( x+1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+1 \right)}
\end{align}$
Now, take the L.C.M. on the right side:
$\frac{6{{x}^{2}}-x+1}{{{x}^{3}}+{{x}^{2}}+x+1}=\frac{A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}$.
By eliminating the denominators from both sides, we get:
$\begin{align}
& 6{{x}^{2}}-x+1=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right) \\
& =A{{x}^{2}}+A+B{{x}^{2}}+Bx+Cx+C \\
& =\left( A+B \right){{x}^{2}}+\left( B+C \right)x+A+C
\end{align}$
Now, compare the coefficients of ${{x}^{2}},x$ and the constant term:
$A+B=6$ …… (I)
$B+C=-1$ …… (II)
$A+C=1$ …… (III)
And find the value of A from equation (I):
$B=6-A$ …… (IV)
And value of B put in the equation (II):
$6-A+C=-1$
$-A+C=-7$ ……m (V)
And add equation (V) with equation (III):
$\begin{align}
& -A+C=-7 \\
& A+C=1 \\
& 2C=-6 \\
& C=-3
\end{align}$
And put the value of C in equation (II):
$\begin{align}
& B-3=-1 \\
& B=2
\end{align}$
And put the value of B in equation (I):
$\begin{align}
& A+2=6 \\
& A=4
\end{align}$
Then,
$\frac{6{{x}^{2}}-x+1}{{{x}^{3}}+{{x}^{2}}+x+1}=\frac{4}{\left( x+1 \right)}+\frac{2x-3}{\left( {{x}^{2}}+1 \right)}$
Thus, the partial fraction of the provided expression is $\frac{4}{\left( x+1 \right)}+\frac{2x-3}{\left( {{x}^{2}}+1 \right)}$.
