Answer
The partial fraction is, $\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$.
$\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+1 \right)}$.
Work Step by Step
Take the L.C.M of the right side:
$\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x-1 \right)}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}$
By eliminating the denominators from both sides, we get:
$\begin{align}
& 5{{x}^{2}}-6x+7=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x-1 \right) \\
& =A{{x}^{2}}+A+B{{x}^{2}}-Bx+Cx-C \\
& =\left( A+B \right){{x}^{2}}+\left( C-B \right)x+A-C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A+B=5$ …… (I)
$C-B=-6$ …… (II)
$A-C=7$ …… (III)
And add equation (II) and equation (III):
$A-B=1$ …… (IV)
Then, add equation (IV) to equation (I):
$\begin{align}
& 2A=6 \\
& A=3
\end{align}$
And the value of A is put in equation (I):
$\begin{align}
& 3+B=5 \\
& B=2
\end{align}$
And the value of B is put in equation (II):
$\begin{align}
& C-2=-6 \\
& C=-4
\end{align}$
Now, $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$.
Thus, the partial fraction of the provided expression is $\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$.
