Answer
The partial fraction is, $\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$.
$\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{A}{\left( x-4 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+5 \right)}$
Work Step by Step
Now, take L.C.M on the right side:
$\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{A\left( {{x}^{2}}+5 \right)+\left( Bx+C \right)\left( x-4 \right)}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}$.
By eliminating the denominators from both sides, we get:
$\begin{align}
& 5{{x}^{2}}-9x+19=A\left( {{x}^{2}}+5 \right)+\left( Bx+C \right)\left( x-4 \right) \\
& =A{{x}^{2}}+5A+B{{x}^{2}}-4Bx+Cx-4C \\
& =\left( A+B \right){{x}^{2}}+\left( C-4B \right)x+5A-4C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and the constant term
$A+B=5$ …… (I)
$C-4B=-9$ …… (II)
$5A-4C=19$ …… (III)
Equation (I): multiply by 4 and add with equation (II):
$\begin{align}
& 4A+4B=20 \\
& C-4B=-9
\end{align}$
$4A+C=11$ …… (IV)
Equation (IV) multiplies by 4 and adds with equation (III):
$\begin{align}
& 16A+4C=44 \\
& 5A-4C=19 \\
& 21A=\ 63 \\
& A=3
\end{align}$
Then, put the value of A in equation (IV):
$\begin{align}
& 12+C=11 \\
& C=-1
\end{align}$
Also, put the value of A in equation (I)
$\begin{align}
& 3+B=5 \\
& B=2
\end{align}$
Then,
$\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$
Thus, the partial fraction of the provided expression is $\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$.