Answer
The partial fraction decomposition of the rational expression is $\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)}$
Work Step by Step
By putting a constant $ A $ over the linear factor $ x-2$ and a linear expression $ Bx+C $ over the quadratic factor:
$\frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1}$
And multiply both sides of the equation by $\left( x-2 \right)\left( {{x}^{2}}+1 \right)$ as given below:
$\begin{align}
& \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left( x-2 \right)\left( {{x}^{2}}+1 \right)\left( \frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1} \right) \\
& \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left\{ \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{A}{x-2}+\left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{Bx+C}{{{x}^{2}}+1} \right\}
\end{align}$
And divide out the common factors:
$\begin{align}
& \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\left\{ \left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{A}{\left( x-2 \right)}+\left( x-2 \right)\left( {{x}^{2}}+1 \right)\cdot \frac{Bx+C}{{{x}^{2}}+1} \right\} \\
& 3x=\left\{ \left( {{x}^{2}}+1 \right)\cdot A+\left( x-2 \right)\cdot \left( Bx+C \right) \right\} \\
& 3x=A{{x}^{2}}+A+\left( B{{x}^{2}}-2Bx+Cx-2C \right) \\
& 3x=\left( A+B \right){{x}^{2}}+\left( -2B+C \right)x+\left( A-2C \right)
\end{align}$
Then, equate the coefficients of like terms of the equation to write a system of equations as shown below:
$ A+B=0$ (I)
$-2B+C=3$ (II)
$ A-2C=0$ (III)
From equations (I) and (III), it is obtained that $ A=-B\text{ and }A=2C $.
Put $ A=-B $ in equation (III):
$\begin{align}
& A-2C=0 \\
& -B-2C=0
\end{align}$ (IV)
Multiply by 2 in equation (II) and add with equation (IV):
$\begin{align}
& +\text{ }\underline{\begin{align}
& -4B+2C=6 \\
& -B-2C=0
\end{align}} \\
& \text{ }-5B=6 \\
\end{align}$
Divide both sides by $-5$ as given below:
$\begin{align}
& \left( \frac{-5B}{-5} \right)=\frac{6}{-5} \\
& B=-\frac{6}{5}
\end{align}$
So, the value of $ B $ is $-\frac{6}{5}$.
Then, put the value of $ B $ in equation (IV) to get the value of $ C $:
$\begin{align}
& -B-2C=0 \\
& -\left( -\frac{6}{5} \right)-2C=0 \\
& 2C=-\frac{6}{5} \\
& C=-\frac{6}{10}
\end{align}$
As $ A=-B $, $ A=\frac{6}{5}$
Put the values of $ A,\,\,B,\text{ and }C $ in the given equation, and define the partial fraction decomposition:
$\begin{align}
& \frac{3x}{\left( x-2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{{{x}^{2}}+1} \\
& =\frac{\frac{6}{5}}{\left( x-2 \right)}+\frac{-\frac{6}{5}x-\frac{6}{10}}{{{x}^{2}}+1} \\
& =\frac{6}{5\left( x-2 \right)}+\frac{\frac{-12x-6}{10}}{{{x}^{2}}+1} \\
& =\frac{6}{5\left( x-2 \right)}-\frac{\not{2}\left( 6x+3 \right)}{1\not{0}\left( {{x}^{2}}+1 \right)} \\
& =\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)}
\end{align}$
Thus, the partial fraction decomposition of the rational expression is $\frac{6}{5\left( x-2 \right)}-\frac{\left( 6x+3 \right)}{5\left( {{x}^{2}}+1 \right)}$.
