Answer
$\dfrac{3x}{(x-2)(x^2+1)}=\dfrac{6}{5(x-2)}-\dfrac{6x-3}{5(x^2+1)}$
Work Step by Step
We are given the fraction:
$\dfrac{3x}{(x-2)(x^2+1)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{3x}{(x-2)(x^2+1)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+1}$
Multiply all terms by the least common denominator $(x-2)(x^2+1)$:
$(x-2)(x^2+1)\cdot\dfrac{3x}{(x-2)(x^2+1)}=(x-2)(x^2+1)\cdot\dfrac{A}{x-2}+(x-2)(x^2+1)\cdot\dfrac{Bx+C}{x^2+1}$
$3x=A(x^2+1)+(Bx+C)(x-2)$
$3x=Ax^2+A+Bx^2-2Bx+Cx-2C$
$3x=(A+B)x^2+(-2B+C)x+(A-2C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=0\\
-2B+C=3\\
A-2C=0
\end{cases}$
Solve the system:
$\begin{cases}
2A+2B=2(0)\\
-2B+C=3\\
A-2C=0
\end{cases}$
$\begin{cases}
2A+2B-2B+C=0+3\\
A-2C=0
\end{cases}$
$\begin{cases}
2A+C=3\\
A-2C=0
\end{cases}$
$\begin{cases}
4A+2C=2(3)\\
A-2C=0
\end{cases}$
$4A+2C+A-2C=6+0$
$5A=6$
$A=\dfrac{6}{5}$
$A-2C=0$
$\dfrac{6}{5}-2C=0$
$\dfrac{6}{5}=2C$
$C=\dfrac{3}{5}$
$A+B=0$
$\dfrac{6}{5}+B=0$
$B=-\dfrac{6}{5}$
The partial fraction decomposition is:
$\dfrac{3x}{(x-2)(x^2+1)}=\dfrac{\dfrac{6}{5}}{x-2}+\dfrac{-\dfrac{6}{5}x+\dfrac{3}{5}}{x^2+1}$
$\dfrac{3x}{(x-2)(x^2+1)}=\dfrac{6}{5(x-2)}-\dfrac{6x-3}{5(x^2+1)}$
