Answer
$ \displaystyle \frac{1}{4(x-1)}+\frac{3}{4(x+3)}$
Work Step by Step
First, factor the trinomial in the denominator
(Two factors of $-3$ that add to $+2$ are $3$ and $-1$)
$2x^{2}-5x-3=(x-1)(x+3)$
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{x}{(x-1)(x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+3)}$
multiply both sides with $(x-1)(x+3)\quad $(LCD)
$x=A(x+3)+B(x-1)$
$ x=Ax+3A+Bx-B\qquad$ /group like terms...
$x+0=x(A+B)+(3A-B)$
... we now equate the coefficients of like powers...
$\left\{\begin{array}{ll}
I. & 1=A+B\\
II. & 0=3A-B
\end{array}\right.\qquad $solve the system
$(II+I)$ eliminates $B...$
$1+0=3A+A-B+B$
$1=4A$
$A=\displaystyle \frac{1}{4}$
back substitute: $1=A+B$
$1=\displaystyle \frac{1}{4}+B$
$B=\displaystyle \frac{3}{4}$
Rewrite the setup decomposition with $A=\displaystyle \frac{1}{4}, B=\frac{3}{4}$
$\displaystyle \frac{x}{(x-1)(x+3)}=\frac{\frac{1}{4}}{(x-1)}+\frac{\frac{3}{4}}{(x+3)}$
$\displaystyle \frac{x}{(x-1)(x+3)}= \frac{1}{4(x-1)}+\frac{3}{4(x+3)}$
