## Precalculus (6th Edition) Blitzer

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