## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$. $\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A}{\left( x+1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+2x+2 \right)}$
Now, take the L.C.M on the right side: $\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}$. By eliminating the denominators from both sides, we get: \begin{align} & 5{{x}^{2}}+6x+3=A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x+1 \right) \\ & =A{{x}^{2}}+2Ax+2A+B{{x}^{2}}+Bx+Cx+C \\ & =\left( A+B \right){{x}^{2}}+\left( 2A+C+B \right)x+2A+C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=5$ …… (I) $2A+C+B=6$ …… (II) $2A+C=3$ …… (III) And put the value of equation (III) in equation (II): \begin{align} & 3+B=6 \\ & B=3 \end{align} Also, put the value of B into equation (I): \begin{align} & A+3=5 \\ & A=2 \end{align} Also, substitute the value of A in equation (III): \begin{align} & 4+C=3 \\ & C=-1 \end{align} Then, $\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$. Thus, the partial fraction of the provided expression is $\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$.