Answer
The partial fraction is, $\frac{14}{3\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\left( -14x+4 \right)}{3\left( {{x}^{2}}+2 \right)}$.
Work Step by Step
$\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{{{\left( x-1 \right)}^{2}}}+\frac{Cx+D}{\left( {{x}^{2}}+2 \right)}$.
Now, take the L.C.M. on the right side:
$\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{A\left( x-1 \right)\left( {{x}^{2}}+2 \right)+B\left( {{x}^{2}}+2 \right)+\left( Cx+D \right){{\left( x-1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}$.
By eliminating the denominator from both sides, we get
$\begin{align}
& 10{{x}^{2}}+2x=A\left( x-1 \right)\left( {{x}^{2}}+2 \right)+B\left( {{x}^{2}}+2 \right)+\left( Cx+D \right){{\left( x-1 \right)}^{2}} \\
& =A\left( {{x}^{3}}+2x-{{x}^{2}}-2 \right)+B{{x}^{2}}+2B+\left( Cx+D \right)\left( {{x}^{2}}+1-2x \right) \\
& =A{{x}^{3}}-A{{x}^{2}}+2Ax-2A+B{{x}^{2}}+2B+C{{x}^{3}}+Cx-2C{{x}^{2}}+D{{x}^{2}}+D-2Dx \\
& =\left( A+C \right){{x}^{3}}+\left( -A+B-2C+D \right){{x}^{2}}+\left( 2A+C-2D \right)x-2A+2B+D
\end{align}$
Then, compare the coefficients of ${{x}^{3}},\ {{x}^{2}},{{x}^{1}}$ and the constant term:
$A+C=0$ …… (I)
$-A+B-2C+D=10$ …… (II)
$2A+C-2D=2$ …… (III)
$-2A+2B+D=0$ …… (IV)
Then, find the value of C from equation (I):
$C=-A$ …… (V)
And the value of C is put in equation (III):
$2A-A-2D=2$
$A-2D=2$ …… (VI)
Now, find the value of B from equation (II):
$B=10+A+2C-D$ …… (VII)
Substitute the value of C in equation (VII):
$B=10+A+2\left( -A \right)-D$
$B=10-A-D$ …… (VIII)
Also, put the value of B in equation (IV):
$\begin{align}
& -2A+2\left( 10-A-D \right)+D=0 \\
& -2A+20-2A-2D+D=0
\end{align}$
$4A+D=20$ …… (IX)
Also, multiply equation (IX) by 2 and add with equation (VI):
$\begin{align}
& 8A+2D=40 \\
& A-2D=2 \\
& 9A=42 \\
& A=\frac{42}{9}=\frac{14}{3}
\end{align}$
And put the value of A in equation (I):
$\begin{align}
& \frac{14}{3}+C=0 \\
& C=\frac{-14}{3}
\end{align}$
Also, put the value of A in equation (VI):
$\begin{align}
& \frac{14}{3}-2D=2 \\
& 2D=\frac{14}{3.}-2 \\
& 2D=\frac{14-6}{3} \\
& D=\frac{4}{3}
\end{align}$
Also, substitute the values of A and D in equation (VIII):
$\begin{align}
& B=10-\frac{14}{3}-\frac{4}{3} \\
& =\frac{30-14-4}{3} \\
& =\frac{12}{3} \\
& =4
\end{align}$
Then,
$\frac{10{{x}^{2}}+2x}{{{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+2 \right)}=\frac{\frac{14}{3}}{\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\frac{-14}{3}x+\frac{4}{3}}{\left( {{x}^{2}}+2 \right)}$.
Thus, the partial fraction of the provided expression is $\frac{14}{3\left( x-1 \right)}+\frac{4}{{{\left( x-1 \right)}^{2}}}+\frac{\left( -14x+4 \right)}{3\left( {{x}^{2}}+2 \right)}$.
