## Precalculus (6th Edition) Blitzer

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