Answer
$\dfrac{6x^2-x+1}{(x+1)(x^2+1)}=\dfrac{4}{x+1}+\dfrac{2x-3}{x^2+1}$
Work Step by Step
We are given the fraction:
$\dfrac{6x^2-x+1}{x^3+x^2+x+1}$
Factor the denominator:
$\dfrac{6x^2-x+1}{x^3+x^2+x+1}=\dfrac{6x^2-x+1}{x^2(x+1)+(x+1)}=\dfrac{6x^2-x+1}{(x+1)(x^2+1)}$
We can write the partial fraction decomposition:
$\dfrac{6x^2-x+1}{(x+1)(x^2+1)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+1}$
Multiply all terms by the least common denominator $(x+1)(x^2+1)$:
$(x+1)(x^2+1)\cdot\dfrac{6x^2-x+1}{(x+1)(x^2+1)}=(x+1)(x^2+1)\cdot\dfrac{A}{x+1}+(x+1)(x^2+1)\cdot\dfrac{Bx+C}{x^2+1}$
$6x^2-x+1=A(x^2+1)+(Bx+C)(x+1)$
$6x^2-x+1=Ax^2+A+Bx^2+Bx+Cx+C$
$6x^2-x+1=(A+B)x^2+(B+C)x+(A+C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=6\\
B+C=-1\\
A+C=1
\end{cases}$
Solve the system: multiply Equation 2 by -1 and add it to Equation 3:
$\begin{cases}
A+B=6\\
B+C=-1\\
A+C-B-C=1-(-1)
\end{cases}$
$\begin{cases}
A+B=6\\
A-B=2
\end{cases}$
$A+B+A-B=6+2$
$2A=8$
$A=4$
$A+B=6$
$4+B=6$
$B=2$
$A+C=1$
$4+C=1$
$C=-3$
The partial fraction decomposition is:
$\dfrac{6x^2-x+1}{(x+1)(x^2+1)}=\dfrac{4}{x+1}+\dfrac{2x-3}{x^2+1}$
