#### Answer

$\dfrac{2}{x+2}+\dfrac{1}{x-1} $

#### Work Step by Step

We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor.
So, we will write the given rational expression into partial fraction form as:
$$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\quad \quad(1)$$
Now, write the RHS with a common denominator using their $LCD$.
\begin{align*}
\frac{3x}{(x+2)(x-1)}&=\frac{A(x-1)+B(x+2)}{(x+2) (x-1)}\\
\\
\end{align*}
Equate the numerators to obtain:
$$3x= A(x-1)+B(x+2)
\\ 3x=Ax-A+Bx+2B
\\ 3x=Ax-A+Bx+2B
\\ 3x=Ax+Bx-A+2B
\\3x= (A+B)x+(-A+2B)$$
Equate the coefficients of the polynomials on the $LHS$ and $RHS$ to obtain:
$$A+B=3\quad \quad(2)
\\ -A+2B=0 \quad (3)$$
Add Equations $(2)$ and $(3)$ to obtain:
$$(A+B)+(-A+2B)=3+0
\\3B=3
\\ B=1$$
Substitute $B=1$ into Equation $(3)$ to solve for $A$.
$$-A+(2)(1)=0\\ -A+2=0 \\ A=2$$
Thus, equation $(1)$ becomes:
$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{2}{x+2}+\frac{1}{x-1} $