Answer
$\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}=\dfrac{14}{3(x-1)}+\dfrac{4}{(x-1)^2}-\dfrac{14x-4}{3(x^2+2)}$
Work Step by Step
We are given the fraction:
$\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{Cx+D}{x^2+2}$
Multiply all terms by the least common denominator $(x-1)^2(x^2+2)$:
$(x-1)^2(x^2+2)\cdot\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}=(x-1)^2(x^2+2)\cdot\dfrac{A}{x-1}+(x-1)^2(x^2+2)\cdot\dfrac{B}{(x-1)^2}+(x-1)^2(x^2+2)\cdot\dfrac{Cx+D}{x^2+2}$
$10x^2+2x=A(x-1)(x^2+2)+B(x^2+2)+(Cx+D)(x-1)^2$
$10x^2+2x=A(x^3+2x-x^2-2)+Bx^2+2B+(Cx+D)(x^2-2x+1)$
$10x^2+2x=Ax^3+2Ax-Ax^2-2A+Bx^2+2B+Cx^3-2Cx^2+Cx+Dx^2-2Dx+D$
$10x^2+2x=(A+C)x^3+(-A+B-2C+D)x^2+(2A+C-2D)x+(-2A+2B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+C=0\\
-A+B-2C+D=10\\
2A+C-2D=2\\
-2A+2B+D=0
\end{cases}$
Solve the system: multiply Equation 1 by 2 and add it to Equation 2, multiply Equation 1 by -1 and add it to Equation 3:
$\begin{cases}
-A+B-2C+D+2A+2C=10+2(0)\\
2A+C-2D-A-C=2-0\\
-2A+2B+D=0
\end{cases}$
$\begin{cases}
A+B+D=10\\
A-2D=2\\
-2A+2B+D=0
\end{cases}$
Multiply Equation 1 by 2 and add it to Equation 2, multiply Equation 1 by -1 and add it to Equation 3:
$\begin{cases}
A-2D+2A+2B+2D=2+2(10)\\
-2A+2B+D-A-B-D=0-10
\end{cases}$
$\begin{cases}
3A+2B=22\\
-3A+B=-10
\end{cases}$
$3A+2B-3A+B=22-10$
$3B=12$
$B=4$
$-3A+B=-10$
$-3A+4=-10$
$-3A=-14$
$A=\dfrac{14}{3}$
$A-2D=2$
$\dfrac{14}{3}-2D=2$
$2D=\dfrac{14}{3}-2$
$2D=\dfrac{8}{3}$
$D=\dfrac{4}{3}$
$A+C=0$
$\dfrac{14}{3}+C=0$
$C=-\dfrac{14}{3}$
The partial fraction decomposition is:
$\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}=\dfrac{\dfrac{14}{3}}{x-1}+\dfrac{4}{(x-1)^2}+\dfrac{-\dfrac{14}{3}x+\dfrac{4}{3}}{x^2+2}$
$\dfrac{10x^2+2x}{(x-1)^2(x^2+2)}=\dfrac{14}{3(x-1)}+\dfrac{4}{(x-1)^2}-\dfrac{14x-4}{3(x^2+2)}$
