Answer
$\dfrac{x}{(x+3)(x-1)}=\dfrac{\frac{3}{4}}{x+3}+\dfrac{\frac{1}{4}}{x-1} $
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has non-repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{x}{(x+3)(x-1)}=\frac{A}{x+3}+\frac{B}{x-1} ~~~(1)$$
Now, write the RHS with a common denominator $$\displaystyle \frac{x}{(x+3)(x-1)}=\dfrac{A(x-1)+B(x+3)}{(x+3)(x-1)}\\ \displaystyle \frac{x}{(x+3)(x-1)}=\dfrac{Ax-A+Bx+3B}{(x+3)(x-1)} \\ \displaystyle \frac{x}{(x+3)(x-1)}=\dfrac{(A+B)x+(-A+3B)}{(x+3)(x-1)}$$
Next, equate the numerators to obtain:
$$ (A+B)x+(-A+3B)=x$$
Equate the like coefficients of the polynomials on the LHS and RHS to obtain: $$A+B=1 ~~~(2) \\ -A+3B=0 ~~~(3)$$
Equation (2) yields: $B=1-A ~~~~~~~(4)$
Substitute $B=1-A$ into equation (3) to solve for $A$. $$-A+3(1-A) =0\\ -A+3-3A=0\\-4A+3=0\\-4A=-3\\ A=\dfrac{3}{4}$$ Finally, substitute $A=\dfrac{3}{4}$ into equation (4) to solve for $B$. $$B=1-\dfrac{3}{4}\\ B=\dfrac{4-3}{4}\\ B=\dfrac{1}{4}$$
Therefore, the equation (1) becomes:
$\dfrac{x}{(x+3)(x-1)}=\dfrac{\frac{3}{4}}{x+3}+\dfrac{\frac{1}{4}}{x-1} $
