## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$. $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+1 \right)}$.
Take the L.C.M of the right side: $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x-1 \right)}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}$ By eliminating the denominators from both sides, we get: \begin{align} & 5{{x}^{2}}-6x+7=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x-1 \right) \\ & =A{{x}^{2}}+A+B{{x}^{2}}-Bx+Cx-C \\ & =\left( A+B \right){{x}^{2}}+\left( C-B \right)x+A-C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=5$ …… (I) $C-B=-6$ …… (II) $A-C=7$ …… (III) And add equation (II) and equation (III): $A-B=1$ …… (IV) Then, add equation (IV) to equation (I): \begin{align} & 2A=6 \\ & A=3 \end{align} And the value of A is put in equation (I): \begin{align} & 3+B=5 \\ & B=2 \end{align} And the value of B is put in equation (II): \begin{align} & C-2=-6 \\ & C=-4 \end{align} Now, $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$. Thus, the partial fraction of the provided expression is $\frac{3}{\left( x-1 \right)}+\frac{2x-4}{\left( {{x}^{2}}+1 \right)}$.