Answer
$\dfrac{5x^2-6x+7}{(x-1)(x^2+1)}=\dfrac{3}{x-1}+\dfrac{2x-4}{x^2+1}$
Work Step by Step
We are given the fraction:
$\dfrac{5x^2-6x+7}{(x-1)(x^2+1)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{5x^2-6x+7}{(x-1)(x^2+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}$
Multiply all terms by the least common denominator $(x-1)(x^2+1)$:
$(x-1)(x^2+1)\cdot\dfrac{5x^2-6x+7}{(x-1)(x^2+1)}=(x-1)(x^2+1)\cdot\dfrac{A}{x-1}+(x-1)(x^2+1)\cdot\dfrac{Bx+C}{x^2+1}$
$5x^2-6x+7=A(x^2+1)+(Bx+C)(x-1)$
$5x^2-6x+7=Ax^2+A+Bx^2-Bx+Cx-C$
$5x^2-6x+7=(A+B)x^2+(-B+C)x+(A-C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=5\\
-B+C=-6\\
A-C=7
\end{cases}$
Solve the system: add Equation 2 to Equation 3:
$\begin{cases}
A+B=5\\
-B+C+A-C=-6+7
\end{cases}$
$\begin{cases}
A+B=5\\
A-B=1
\end{cases}$
$A+B+A-B=5+1$
$2A=6$
$A=3$
$A-C=7$
$3-C=7$
$C=-4$
The partial fraction decomposition is:
$\dfrac{5x^2-6x+7}{(x-1)(x^2+1)}=\dfrac{3}{x-1}+\dfrac{2x-4}{x^2+1}$
