Answer
The partial fraction is, $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
Work Step by Step
$\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right)}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$
Now, take the L.C.M. on the right side:
$\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$
By eliminating the denominator from both sides, we get
$\begin{align}
& {{x}^{2}}+2x+3=\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right) \\
& =A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+Cx+D \\
& =A{{x}^{3}}+B{{x}^{2}}+\left( 4A+C \right)x+4B+D
\end{align}$
Then compare the coefficients of ${{x}^{3}},\ {{x}^{2}},{{x}^{1}}$ and the constant term:
$A=0$ …… (I)
$B=1$ …… (II)
$4A+C=2$ …… (III)
$4B+D=3$ …… (IV)
Substitute the value of A in equation (III):
$C=2$
And put the value of B in equation (IV):
$\begin{align}
& 4+D=3 \\
& D=-1
\end{align}$
Now,
$\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
Thus, the partial fraction of the given expression is $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
