Answer
The partial fraction function is $\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$.
Work Step by Step
Let us consider the provided expression,
$y=\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$ …… (I)
Then, find the partial fraction as given below,
$\begin{align}
& \frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}=\frac{A}{\left( {{x}^{2}}+7 \right)}+\frac{C}{{{\left( {{x}^{2}}+7 \right)}^{2}}} \\
& {{\left( {{x}^{2}}+7 \right)}^{2}}\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}={{\left( {{x}^{2}}+7 \right)}^{2}}\left( \frac{A}{\left( {{x}^{2}}+7 \right)}+\frac{C}{{{\left( {{x}^{2}}+7 \right)}^{2}}} \right) \\
& 7{{x}^{2}}-9x+3=A\left( {{x}^{2}}+7 \right)+C \\
& 7{{x}^{2}}-9x+3=A{{x}^{2}}+7A+C
\end{align}$ …… (II)
Now, compare the coefficient of the equation (II) of ${{x}^{2}},{{x}^{1}}$ and ${{x}^{0}}$:
$A=7$ …… (III)
$7A+C=3$ …… (IV)
And put the value of equation (III) into equation (IV),
$\begin{align}
& 7\left( 7 \right)+C=3 \\
& C=3-49 \\
& =-46
\end{align}$
Therefore,
$\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}=\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$
Thus, the partial fraction of the given expression is $\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$.
