In Problems 13–46, write the partial fraction decomposition of each rational expression.

$\frac{10x^2+2x}{{(x-1)}^2(x^2+2)}$

Given rational expression is $\frac{10x^2+2x}{{(x-1)}^2(x^2+2)}$.

We have to write the partial fraction decomposition of the given rational expression.

We decompose the given rational expression as follows:

$\frac{10x^2+2x}{{(x-1)}^2(x^2+2)}=\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{Cx+D}{(x^2+2)}$

We have to find $A,B,C,D$, so multiply both sides with ${(x-1)}^2(x^2-2)$.

$10x^2+2x=A(x-1)(x^2+2)+B(x^2+2)+(Cx+D){(x-1)}^2$ - Equation-2

Substitute $x=1$in Equation-2

$$\begin{gathered} 10+2=3B \\ \Rightarrow B=\frac{12}{3} \\ \Rightarrow B=4 \end{gathered}$$

• Substitute $x=0$in Equation-2.

role="math" localid="1647461457889" $$\begin{gathered} 0=A(-1)2+B\left(2\right)+D{(-1)}^2 \\ \Rightarrow 2A-2B-D=0 \end{gathered}$$

Substitute value of B.

$2A-D=8$ -Equation-3

• Substitute $x=-1$in Equation-2

$$\begin{gathered} 10-2=A(-2)\left(3\right)+B\left(3\right)+(-C+D)4 \\ \Rightarrow 6A+4C-4D+8=3B \end{gathered}$$

Substitute value of B.

role="math" localid="1647461805431" $$\begin{gathered} 6A+4C-4D+8=12 \\ \Rightarrow 6A+4C-4D=4 \end{gathered}$$

$3A+2C-2D=2$ - Equation-4

• Substitute $x=2$in Equation-2

$$\begin{gathered} 10\left(4\right)+2\left(2\right)=6A+6B+C+D \\ \Rightarrow 6A+C+D-44=-6B \end{gathered}$$

Substitute value of B.

$$\begin{gathered} 6A+C+D-44=-24 \\ \Rightarrow 6A+C+D=20 \end{gathered}$$

$6A+C+D=20$ -Equation-5

Solving the equations 3,4,5 will give:'

$$\begin{gathered} A=\frac{14}{3} \\ C=-\frac{14}{3} \\ D=\frac{4}{3} \end{gathered}$$

Therefore by Substituting the value of $A,B,C,D$the partial fraction decomposition is:

$\frac{10x^2+2x}{{(x-1)}^2(x^2+2)}=\frac{14}{3(x-1)}+\frac{4}{{(x-1)}^2}+\frac{-14x+4}{3(x^2+2)}$