Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
$\frac{{{x}^{2}}+2x+3}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right)}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$ 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 $\frac{1}{\left( {{x}^{2}}+4 \right)}+\frac{2x-1}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.